1 Light reflected by interface

A mixture of red light (λvacuum = 661 nm) and blue light (λvacuum = 472 nm)shines perpendicularly on a thin layer of gasoline (ngas = 1.40 lying on water(nwater = 1.33). The gasoline layer has a uniform thickness of 1.69 × 10−7 m.What color is the light reflected from the gasoline? Neglect any effects(s) causedby the material under the water.

Solution

Because the light wave reflected from the air/gas interface is inverted, and that fromthe gas/water interface is not (all transmitted waves are transmitted in phase), we willget destructive interference when the path difference of the two waves in the gasolineis an integral number of wavelengths, and constructive interference when it is a half-integral number of wavelengths. The path difference is always twice the thickness ofthe gasoline layer for normal incidence. The wavelength of light in a medium is

λ =c/n

ν=λvacuum

n,

since the time frequency of a transmitted wave cannot change (think of a particle onthe boundary responding to the incoming EM wave) and the speed of light in themedium is reduced by dividing by n. This means that the number of wavelengths oflight in the path difference δ = 2t is

N =2tn

λvacuum,

which is 0.716 for the red light and 1.00 for the blue light. This means that the

interference for the blue light is destructive and we will see predominantly red light.

Note that since the intensities of the two reflected blue waves will not exactly match

because of the different ratios of the indices of refraction at the two boundaries, we

will not get a perfect cancellation of the reflected blue light.

Problem OpaTh-F99-12 from Courses-SC.

2 Specific heat of magnetic solid

A solid contains N magnetic atoms having spin 12 . At sufficiently high tem-

perature, each spin is completely randomly oriented, i.e. equally likely to bein either of its two possible states. But at sufficiently low temperature, the in-teraction between the magnetic atoms causes them to exhibit ferromagnetism,with the result that all their spins become oriented along the same direction asT → 0. A very crude approximation suggests that the spin-dependent contri-bution C(T ) to the heat capacity of this solid has an approximate temperaturedependence given by

C(T ) =

C1

(2TT1− 1)

if T1/2 < T < T1

0 otherwise

Use entropy considerations to find the maximum possible value of C1.

Solution

If we know the specific heat we can find the entropy by using the specific heat tofind the heat added, and then using that the entropy is related to the heat added bydQ = TdS. For a system of spins we can also find the entropy using the number ofmicrostates available to the system. If we relate the entropies found in these two wayswe can find a relation for C1.If dQ = TdS and dQ = C(T )dT , then we have that the relationship between entropyand specific heat is that

dS =C(T )

TdT,

so that we can integrate to find the entropy S

S(T ′) =

∫ T ′

0

C(T )

TdT.

If we choose a final temperature T ′ large compared to T1, then we have

S(T ′) =

∫ T ′

0

C(T )

TdT

=

∫ T1

T1/2

C1

(2

T1− 1

T

)dT

= C1

[2T

T1− ln(T )

]T1

T1/2

= C1 [1− ln(2)] .

Now when the system is at very low temperature the spins are all aligned and thenumber of microstates available to the system is very small–just 2! This means theentropy is simply

S = kB ln(Ω) = kB ln(2)

Problem OpaTh-F00-2 from Courses-SC..

at very low temperature. As we raise the temperature a few of the spins will be ableto be misaligned, until at very high temperature all spin configurations (including halfof the spins aligned one way and half the other way, for example, and the states whereall the spins are aligned) are possible. If the spins are aranged in a regular lattice andcan be labeled, then the largest possible number of available microstates of the systemis given by two choices for spin 1, times two for spin 2, etc, for a total of Ω = 2N , sothat the maximum value of the entropy is

S = kB ln(Ω) ≤ kB ln(2N ) = NkB ln(2).

Comparing this to our result above, which is also for high temperature, we have

C1 [1− ln(2)] ≤ NkB ln(2),

so that

C1 ≤ NkBln(2)

1− ln(2).

3 Two converging lenses

Two converging thin lenses, L1 and L2, each of focal length f1 = f2 = 10 cm,are separated by the distance d = 35 cm. An object (the upright arrow) isplaced a distance O1 = 20 cm to the left of the left-hand lens (L1). See thefigure. The check marks on the horizontal axis represent the focal points of thetwo lenses.

(a) Draw a ray diagram, and also find the position of the final image using thethin-lens equation.

(b) Is the final image real or virtual?

(c) Is the final image upright or inverted?

(d) What is the magnification of the final image?

Solution

(a) Since the two focal lengths are both the same, call them both f . Then we canuse the thin lens equation 1/s + 1/s′ = 1/f to find the exact positions of thetwo images, which are shown in the following ray diagram. Note we have usedthe two rays from the tip of the object through the focal point (which comesout parallel) and through the center (which is undeflected) of the lens to locatethe tip of the image, and the base of the image is located by a ray along theaxis of the two lenses.

From the lens equation, using the distances labeled on the figure, we have

1/s′1 = 1/f − 1/s1 = (1/10− 1/20) cm−1,

so that s′1 = 20 cm. This means that s2 = 35− 20 = 15 cm, and so

1/s′2 = (1/10− 1/15) cm−1 =

(3− 2

30

)cm−1 = 1/30 cm−1,

so that s′2 = 30 cm.

Problem OpaTh-F00-2 from Courses-SC.

(b) Obviously the final image is real, as rays of light converge to a point at the imagewhere we can hold a screen.

(c) The final image is obviously upright.

(d) For this we need to calculate the size of the intermediate image from the magnifi-cation m1 = −s′1/s1 = −20/20 = −1, so we see this image is of the same size asthe object but inverted. The second image has magnification m2 = −s′2/s2 =−30/15 = −2, so the combined magnification is m = m1 ·m2 = 2.

4 Ideal gas in confining potential

An ideal classical gas made of N point-like atoms with mass m is placed ina confining potential: V (x, y, z) = 1

2 (kxx2 + kyy

2 + kzz2). The temperature of

the gas is T and there is no external container.

(a) Calculate the average kinetic energy of each gas atom.

(b) Calculate the average potential energy of each gas atom.

(c) Calculate the constant volume heat capacity of the gas.

(d) At low temperatures, the gas has to be considered as composed of quantummechanical particles, and the fact that their energy levels are discreteis important. Briefly discuss what happens to the specific heat at lowtemperatures (no calculation is required).

Solution

(a) The equipartition theorem states that for every independent term quadratic inthe dynamical variables (which are momenta and positions) in the Hamiltonian(energy) of the system we have an average energy of kT/2. The kinetic energyof each gas atom has three such terms, which are p2

x/2m, p2y/2m, and p2

z/2m.Each term has associated with it an average energy of kT/2, so the averagekinetic energy of each gas atom is

〈 p2i

2m〉 =

3

2kT.

(b) Exactly the same argument holds for the potential energy of each particle, whichcontains terms quadratic in the independent position variables x, y, and z. sothat for any i

〈V (ri)〉 =3

2kT.

(c) For an ideal gas heated at constant volume, which can do no work, the first law ofthermodynamics tells us that the heat capacity CV is derivative of the internalenergy U of the gas with temperature T . The internal energy is the sum ofthe average kinetic and potential energies of each gas particle multiplied by thenumber of such particles,

U = N

(3

2kT +

3

2kT

)= 3NkT,

and so

CV =dU

dT= 3Nk.

(d) The third law of thermodynamics states that the entropy S goes to a constantS0 as the temperature goes to zero,

limT→0

S = S0.

Problem OpaTh-S00 from Courses-SC.

We can understand this microscopically by the number of available microstatesof the system tending to a constant at absolute zero, and the entropy is relatedto the natural log of the number of available microstates. Now since the heatcapacity is defined to be

CV =

(dQ

dT

)V

and since dQ = T dS, we have

CV = T

(dS

dT

)V

.

But S approaches a constant as T → 0, so we have

limT→0

CV = 0.

Although somewhat beyond the scope of this question, we can establish howthe specific heat tends to zero. Examine the energy of a single particle in aone-dimensional harmonic potential. The partition function is

Z =∑α

e−βEα ,

where β = 1/kT and the energy eigenvalues Eα are simply (n+ 1/2)hω, whereω is the harmonic oscillator frequency. For this case the partition function isjust a geometric series with the sum

Z =

∞∑n=0

e−(n+1/2)βhω = e−βhω/2∞∑n=0

e−nβhω =e−βhω/2

1− e−βhω .

Now the average energy of the system is

〈E〉 =1

Z

∑α

Eαe−βEα =

1

Z

(−∂Z∂β

)= −∂ lnZ

∂β.

Now we have that

ln(Z) = −1

2βhω − ln(1− e−βhω),

so that

U = 〈E〉 = −∂ lnZ

∂β=

1

2hω +

hωe−βhω

1− e−βhω = hω

(1

2+

1

eβhω − 1

).

The first temperature-independent term is the zero-point energy of the oscillator.The generalization to N particles in three dimensions is simple–each dimensionin which each particle moves behaves like an independent one-dimensional os-cillator, so that the energy is just multiplied by 3N ,

U = 〈E〉3D = 3Nhω

(1

2+

1

eβhω − 1

),

Here we assume kx = ky = kz for simplicity, and now by a straight-forwarddifferentiation we have that

CV = 3Nhω

kT 2

[eβhωhω

(eβhω − 1)2

],

and as the temperature goes to zero we see that

limT→0

CV = 3Nk

(hω

kT

)2

e−hωkT .

We see that, since the exponential goes to zero faster than any power of T , thespecific heat goes to zero exponentially. This is a characteristic of any systemwith an energy gap such as this one (which has an energy gap of hω betweenthe ground state and first excited state). Note the high temperature limit of Cvproduces the ideal gas Cv = 3Nk.

5 Expansion of ideal gas

Consider a reversible isothermal expansion of an ideal gas (the system) withN particles in contact with a heat reservoir at a temperature T , from an initialvolume V1 to a final volume V2.

(a) What is the change of the internal energy of the system?

(b) Obtain the amount of work performed by the system.

(c) What is the amount of heat absorbed by the system?

(d) What is the change of entropy of the system?

(e) What is the change of the entropy of the system plus the reservoir?

Next consider a free expansion of the above ideal gas from an initial volumeV1 to a final volume V2.

(f) What is the change of temperature of the system?

(g) Does the internal energy change?

(h) Obtain the amount of work performed by the system.

(i) Obtain the absorbed amount of heat.

(j) Obtain the change of entropy of the system and that of the reservoir.

Solution

(a) The internal energy of an ideal gas U = CV T = NcV T , where cV is the specificheat per particle and CV is the total specific heat, depends only on the number ofgas molecules and the temperature of the system, so in an isothermal expansionwhere the temperature (and amount of gas) remains the same, the internalenergy does not change, i.e. ∆U = 0.

(b) The work done is

W =

∫ V2

V1

PdV,

and since for an ideal gas at a fixed temperature T we have P = NkBT/V wecan write

W = NkBT

∫ V2

V1

dV

V= NkBT ln

(V2

V1

).

(c) Using the first law of thermodynamics, Q = ∆U +W , and since ∆U = 0 we have

Q = W = NkBT ln

(V2

V1

).

Problem OpaTh-FS01 from Courses-SC.

(d) The change in entropy is

∆S =

∫dS =

∫dQres

T=

1

T

∫dQres =

∆Qres

T,

since the temperature T does not change, and so

∆S = NkB ln

(V2

V1

).

(e) Since the temperature of the heat reservoir does not change by assumption, thensince it loses heat Q to the gas we know that its entropy change is just

∆Sres =Qres

T=−QT

= −NkB ln

(V2

V1

).

Note that this means that the total entropy change is zero, which is the definitionof a reversible process.

(f) In a free expansion, such as a vessel of volume V1 separated from one of volume V2

by a membrane which ruptures, the ideal (and so non-interacting) gas moleculeshave the same kinetic energy before and after the membrane ruptures, so thatthe temperature remains the same.

(g) No, since it depends only on the temperature and that has not changed. Alter-natively, if you think of the motion of each of the molecules before and after themembrane ruptures, the kinetic energy of the molecules has not changed and soneither has U .

(h) When a membrane ruptures, even though V has changed, there is no work doneby the system on its surroundings.

(i) By the first law of thermodynamics, if there is no work done and no change in theinternal energy there can be no heat absorbed.

(j) If there is no heat absorbed from it, then there can be no change in the entropy ofthe reservoir, ∆Sres = 0. The gas, however, has now more available microstates(it has the same total energy but the energy levels are now closer together) andso has more entropy. Another way of saying this is that the entropy of the gasis a function of the state of the gas, and if it has the same state as after theisothermal expansion in Part 1 (d), then it has the same entropy, and so onceagain

∆S = NkB ln

(V2

V1

).

Then the total entropy of the system has increased, as expected for an irre-versible process.

6 Diameters of wires

The diameters of fine wires can be accurately measured using interferencepatterns. Two optically flat pieces of glass of length L are arranged with a wirebetween them as shown in the figure. The setup is illuminated with normallyincident yellow light of wavelength λ = 590 nm. The length is L = 20 cm, and19 bright fringes are seen along this length.

(a) Find the diameter of the wire.

(b) If the 19th fringe is not quite at the end, but there is no 20th fringe, give anestimate of the possible relative error in the measurement of the diameter.

Solution

(a) Where the two plates touch, the light which reflects from the bottom of the toppiece of glass and the light which reflects from the top of the bottom piece ofglass travel have the same path length. However, the light from the secondreflection has its phase inverted by π since it goes from a region of low densityto one of high density, while the opposite is true for the second reflection so itis in phase. That means that these two reflected rays negatively interfere andthere is a dark fringe, as shown. As we move along the top plate, the pathlength difference must increase by λ/2 (for a phase difference of π) by the timewe get to the first bright fringe, and then to λ (for a phase difference of 2π)by the time we get to the second dark fringe, and 3λ/2 for the second brightfringe, and so on. By the time we get to the nineteenth bright fringe, we musthave a path length difference of (18 + 1/2)λ = 37λ/2 when we get to the 19th

bright fringe. Since the path difference is twice the separation of the platesand so twice the diameter of the wire (with the excellent approximation thatthe angular separation of the plates is very small so we have almost normalincidence), then

D = 37λ/4 = 5.47× 10−6 m.

(b) The dark fringe spans a path length difference of λ/2, so we can make a relativeerror of a part in 37, or 2.7%.

Problem OpaTh-FS01-2 from Courses-SC.

7 Otto cycle 2

The operation of a gasoline engine is approximately represented by the Ottocycle. This cycle consists of four different processes:

A→ B The gas is compressed adiabatically from volume VA to volume VB (com-pression stroke).

B → C The gas is heated at constant volume VB (combustion of the gasolinemixture).

C → D The gas expands adiabatically from volume VB to volume VA (powerstroke).

D → A The gas is cooled at constant volume VA (exhaust stroke).

(a) Compute the thermodynamic efficiency of the Otto cycle as a function of thecompression ratio VA/VB and the heat capacity per particle at constantvolume, cV . Assume that the gas is ideal with temperature-independentheat capacities.

(b) Do higher or lower compression ratios give higher efficiency?

Solution

(a) During the constant-volume heating and cooling, by the first law of thermody-

Problem OpaTh-FS01-3 from Courses-SC.

namics (and since no work can be done by or on the gas) we have that

∆Q = ∆U =

∫ 2

1

dU = NCV

∫ T2

T1

dT = NCV (T2 − T1),

so that we have

Qout = NCV (TD − TA), Qin = NCV (TC − TB),

where these are the (positive) heats that are given up and put into the system,respectively. The thermodynamic efficiency of a cycle is defined to be the workdone by the gas divided by the heat put in,

ε =W

|Qin|,

and since the gas does not change its state over an entire cycle we know that itsinternal energy does not change. The first law of thermodynamics then tells usthat the net heat added to the system is the work done by the system, so thatW = |Qin| − |Qout| and the efficiency is

ε =W

|Qin|=|Qin| − |Qout||Qin|

= 1− |Qout||Qin|

=

(1− TD − TA

TC − TB

).

In an adiabatic expansion or compression we have the result (you should knowhow to prove this!) that

TV γ−1 = constant,

where γ = CP /CV is the ratio of the specific heat at constant pressure to thatat constant volume. This will allow us to relate the (unknown) temperatures tothe (known) volumes and so compute the efficiency.

Note thatTCV

γ−1C = TDV

γ−1D ,

so

TC = TD

(VDVC

)γ−1

, TB = TA

(VAVB

)γ−1

,

From the diagram we see that VD/VC = VA/VB which is the compression ratior. Then we have that

ε =

(1− TD − TA

TC − TB

)=

(1− TD − TA

(TD − TA)rγ−1

)=(1− r1−γ) .

Now for an ideal gas we have that CP and CV differ by nR = NkB (this can bederived from the first law of thermodynamics), so

γ − 1 =nR

CV=NkBNcV

=kBcV,

since R = NAkB and nNA = N , and so, finally,

ε = 1− r−kB/cV .

(b) Because γ−1 = kB/cV is a positive definite number for any gas, ε = 1−r1−γ canonly get larger as r gets larger, meaning that higher compression ratios alwaysgive higher efficiencies.

8 Light through film

A loop of wire is dipped into soapy water and forms a soap film when removed.The loop is held so the film is vertical. The index of refraction of the film is 1.4and the light incident on the film has a wavelength of 560 nm in vacuum.

(a) What is the wavelength of the light inside the film?

(b) If the thickness of the film is t = 10−6m, find the number of wavelengthscontained in a thickness 2t.

(c) What happens to the phase of a light wave when it is reflected fromi) the front of the film, andii) the back of the film?

(d) As time progresses, the film becomes thinner due to evaporation. The filmthen appears black in reflected light. Briefly explain why this occurs.

Solution

(a) Recall that the the speed of light in a medium with refractive index n is c/n. Thenνλ = c/n and since ν can’t change (the time taken for one oscillation cannot bealtered by a medium or waves would pile up) then λ = λvacuum/n = 400 nm.

(b) 200× 10−6/(400× 10−9) = 5.

(c) In going from a less dense to a more dense medium (the front of the film), thereflected light has its phase inverted (a phase difference of π) relative to theincident light. In going from a more dense medium to a less dense medium (therear of the film) the reflected light is in phase with the incident light.

(d) The phase difference between the light reflected from the rear of the film and thefront of the film is

δ = 2π

(2t

λ

)+ π,

and as t→ 0 the two reflections destructively interfere.

Problem OpaTh-FS01-4 from Courses-SC.

9 Numerical values

Give approximate numerical values for the following (be sure to give units):

(a) The binding energy of a nucleon in a typical nucleus.

(b) The binding energy of an electron in a hydrogen atom.

(c) The lifetime of a free neutron.

(d) The specific heat of Cu at room temperature (assumed greater than theDebye temp.)

(e) The energy of a photon of light within the visible part of the spectrum.

Solution

(a) The binding energy of a typical nucleon is on the order of ∼ MeV.

(b) The binding energy of an electron within a hydrogen atom is ∼ 13.6 eV (Bohrmodel of the atom).

(c) The lifetime of a free neutron is ∼ 15 minutes. This is roughly the time it takesfor the conversion of a down quark to an up quark through the weak interaction.

(d) The specific heat of Cu at room temperature is ∼ 3R (Dulong-Petit law)

(e) The energy of a photon of light within the visible part of the EM spectrum is ∼eV (E = hf = hc/λ, with λ ∼ 600 nm).

Problem 2 from Qualifying exam 2007-Fall.

10 Reaction and decay

Consider the reaction

p+ p→ p+ p+ Λ + Λ (1)

Assume that the masses of proton and Λ are 1 GeV/c2.

(a) Consider a laboratory frame in which one of the two initial state protonsis at rest. What is the ”threshold” energy, i.e. the minimum energythat the incident proton must have for the reaction to be kinematicallypossible? (Hint: What does this mean in terms of final state momenta inthe center-of-mass frame?)

(b) For the reaction at threshold as described in part (a), what is the meandistance that the Λ travels before it decays? (the mean lifetime of the Λin its rest frame is τ ≈ 2.6× 10−10s)

(c) What is the probability that at least one of the two Λ’s travels the distancedetermined in (b)?

Solution

(Using units such that c = 1.) Let m, Ei, pi be the mass, energy, momentum of theincident proton and pΛ, EΛ the momentum and energy of the Λ, both in the laboratoryframe. Let (Ecms, pcms) be the total energy of the system in the center-of-mass frame,and (Elab, plab) be the total energy of the system in the laboratory frame.

(a) By definition, at threshold the initial energy is just enough to produce the massenergy of the new particles in the final state, i.e. all final state momenta inthe center-of-mass frame are 0. Therefore the total energy in the center-of-massframe is:

Ecms =∑

mi = 4m and pcms = 0. (2)

The total energy and momentum in the laboratory frame is = energy/momentumof incident proton (Ei, pi) + energy/momentum of target proton (m, 0):

Elab = Ei +m =√p2i +m2 +m, plab = pi. (3)

To get the relation between Ecms and the lab energy, we use energy momentumconservation and the fact that the square of the 4-momentum is Lorenz=invariant.

E2cms − p2

cms = E2lab − p2

lab (4)

Since

E2cms − p2

cms = (4m)2 (5)

E2lab − p2

lab = 2m(Ei +m) (6)

Problem 5 from Qualifying exam 2007-spring.

we therefore have (4m)2 = 2m(Ei + m) and thus we get for the energy andmomentum of the incident proton at threshold:

Ei = 7m (7)

and

pi =√E2i −m2 = 4m

√3 (8)

(b) Since in the center-of-mass frame all final state particles have 0 momentum, andsince by assumption they all have the same rest mass, they also have equalmomentum in the lab, i.e. 4pΛ = pi or

pΛ = pi/4 = m√

3, EΛ =√p2

Λ +m2 = 2m. (9)

To calculate the mean distance traveled, we need the speed of Λ. We use therelation

E = mγ, p = mβγ (10)

where

β =v

c, γ =

1√1− β2

. (11)

The speed of Λ is given by c · βΛ, where

βΛ =pΛ

EΛ=

√3

2. (12)

Because of time dilation, the mean lifetime of the Λ’s in the lab τlab = γ · τ .The distance traveled by Λ in this time is d = v · τlab,

d =

√3

2· 3 · 108 · 2 · 2.6 · 10−10m ≈ 13.5cm (13)

(c) The decay follows an exponential decay law: the number N(t) not yet decayedat time t is N(t) = N0e

−t/τ . In other words, the survival probability, i.e. theprobability that a particle’s time of decay tdecay is longer than some given timetd is

P (tdecay ≥ td) = e−td/tau (14)

For td = τ , this gives e−1 ≈ 0.368. Thus the probability that Λ (1) has a lifetimelonger than its mean lifetime is P (1) = 1/e, and for Λ (2) it is P (2) = 1/e. Sincethe decays of the two Λ particles are independent events, the probability for atleast one of the two (either of the two) surviving beyond its mean lifetime is

P (1 or 2) = P (1) + P (2)− P (1 and 2). (15)

The reason we have to subtract P (1 and 2) is to avoid double counting, sinceP (1) includes those cases where in addition to particle (1) also particle (2) livedbeyond its mean time.

Thus the probability that at least one of the two Λ’s will travel the distancedetermined in (b) is

P (at least one) = 2e−1 − e−2 ≈ 0.60. (16)

1 Two nucleon system

The nucleons, proton and neutron, are spin 1/2 particles. Consider a two-nucleon system with a relative angular momentum L and with a total spin S.

(a) For each of the following values of the total angular momentum (J = L+S)and parity: 0+, 0−, 1+, 1−, 2+, and 2−, list all possible combinations ofL and S.

(b) From part (a) list the configurations (L, S, J) that are allowed for identicalparticles.

(c) The interaction between proton and neutron is identical to that betweenthe two neutrons and the potential is very shallow. Given that only protonand neutron have a bound state (called deuteron) and the neutronneutronsystem is unbound, what can you say about the spin and parity of thedeuteron? Justify your answer.

Solution

(a) Let us use the spectroscopic notation 2S+1LJ . Below is the list of allowed config-urations:

0+ 1S0

0− 3P0

1+ 3S1,3D1

1− 1P1,3P1

2+ 1D2,3D2

2− 3P2,3F2.

(b) For identical particles the total wave function must be antisymmetric. The spinpart is symmetric for S = 1 and antisymmetric for S = 0. The orbital part issymmetric if L is even and antisymmetric if L is odd. Thus, the following statesare allowed for identical particles: 1S0, 3P0, 3P1, 1D2, 3P2, 3F2.

(c) The deuteron must be a protonneutron state that is not allowed for identicalparticles; otherwise a bound neutron-neutron system would exist as well. Thismeans deuteron is in 3S1, 3P1, 1P1 or 3D2 as follows from (a) and (b). Due tothe centrifugal potential we must look for a state with the lowest value of theangular momentum L. Therefore deuteron must be in 3S1 state, implying it isa spin-1 parity even bound state, 1+. It can be a mixture of 3S1 and 3D1 butdue to centrifugal barrier it is predominately an S state.

2 Excited nucleus

The excitation energy of a free, excited nucleus at rest in the lab is δE and themass of the excited nucleus is m. The nucleus de-excites by emitting a photon.

(a) Determine the frequency ν of the photon as measured in the lab, accountingfor nuclear recoil. Express your answer in the form

ν = a(δE) + b(δE)2 (1)

and find a and b.

(b) Explain in one sentence the sign of b.

(c) How does your answer change if the nucleus is rigidly attached to a crys-talline lattice? Specifically, find a and b in this case.

Solution

(a) Conservation of momentum gives us-

pN = pν

Momentum of the nucleus, pN is-

pN = mvr

where the subscript r refers to recoil.

Momentum of the photon, pν is-

pν =Eνc

=hν

c

where ν = Eν/h is the frequency of the emiited gamma ray. Hence, we have

mvr =hν

c

Conservation of energy gives us-

δE = hν +mv2r

2

The frequency corresponding to δE (resonant frequency, ν0) is-

δE = hν0

The resonant frequency(in the frame of the nucleus) ν0 is red-shifted to theobserved frequency (lab frame) ν by the recoil of the nucleus (Doppler effect)-

ν = ν0

√c− vrc+ vr

=δE

h

√c− vrc+ vr

Problem ModernPhysics1-2 from 2008Comps.

Hence, we can write the conservation of momentum in terms of the resonantenergy, δE-

mvr =h

c

δE

h

√c− vrc+ vr

v2r = (δE

mc)2c− vrc+ vr

We substitute v2r in the expression for conservation of energy and with sometrivial rearrangement-

ν =δE

h− (

c− vrc+ vr

)(δE)2

2mc2

Comparing with the equation presented to us, we identify a and b as-

a =1

h

b = − 1

2mc2(c− vrc+ vr

)

(b) b is negative, owing to the energy lost in the recoil of the nucleus. The Dopplershift reduces the energy lost by a factor depending on the recoil velocity of thenucleus.

(c) If the nucleus is rigidly attached to a crystalline lattice, the recoil acts on theentire crystal due to the periodic arrangement of atoms in the crystal. Therecoil velocity is significantly killed owing to the large mass of the crystal. Theenergy lost in the recoil process appears as v2r , hence the emitted photon is muchcloser to the resonant frequency. Mathematically, a remains the same, b whichrepresents a measure of the shift from resonance approaches zero.

These calculations form the basis of Mossbauer effect where the recoilless emis-sion of gamma rays from a nucleus embedded in a crystal results in resonantabsorption. Gravitational red shift was verified using Mossbauer spectroscopy.

3 Two rockets

At time T=0,Rocket A leaves the earth with a velocity of c/2 in the x-directionand Rocket B leaves the Earth with a velocity of -2c/3 in the y-direction.Assumethe earth is an inertial frame.Rocket A travels a distance of 6 light years andstops.Call the stopping of Rocket A event 1.Rocket B travels a distance of 4light years relative to the earth’s frame and stops.Call the stopping of RocketB event 2.Both rockets were traveling with constant velocity before stopping.

(a) Are events 1 and 2 time-like,light-like,or space-like separated? You mustexplain your answer to receive credit.

(b) An observer in another reference frame measures a time interval betweenthe event 2 and event 1 of 3 years.What is the distance(in light years)between the event 1 and 2 according to an observer in this frame ?

(c) Consider the time intervals between the event 1 and event 2 recorded byall possible inertial observers.There is an inertial frame that gives theminimum time interval.What is this minimum time interval ?

(d) There is an inertial frame that gives the minimum spatial separation be-tween event 1 and event 2. What is this minimum spatial separation ?

(e) Consider an inertial frame that moves along the direction of the separationsbetween event 1 and event 2 as viewed in the Earth’s reference frame. Findthe velocity vector (relative to Earth) for the frame that gives the shortestpossible distance between events 1 and 2.

Solution

(a)

s2 = (c∆t)2 − x2 − y2 − z2 (2)

∆t = t1 − t2 (3)

t1 =6ly

c/2=

12

cly (4)

t2 =−4ly

−2c/3=

6

cly (5)

∆t =6

cly (6)

s2 = (6ly)2 − (6ly)2 − (4ly)2 = −16ly2 (7)

since s2 is less then zero,events 1 and 2 have to be space-like separated.

Problem MP06 from 2008Comps.

(b) The distance is

∆t′ = 3y (8)

c∆t′ = 3ly (9)

s′ = −16ly2 = (3ly)2 − r2 (10)

r2 = 25ly2 (11)

r = 5ly (12)

(c) Since the two events are space like separated , there must exist a frame where thetwo events occur at the same time but at different locations.Therefore, ∆t′ = 0.

(d) The spatial separation is

s2 = (c∆t)2 − r2 = −16ly2 (13)

∆t′ = 0 (14)

r2 = 16ly2 (15)

r = 4ly (16)

(e)

∆r′ = γ(∆r − βc∆t) (17)

∆t′ = γ(∆t− β

c∆r) (18)

using the assumption that the frame found in parts (c) and (d) travels along thedistance between event 1 and 2

0 = γ(6

cly − β

c

√52ly) (19)

β =6√52

(20)

v =6√52c (21)

4 K decay

Suppose K0 is created with a total energy 10 GeV in the lab.

(a) How far does K0 travel in the lab

(b) Find the smallest possible Kinetic Energy.

(c) Find the angle of decay.

Solution

(a) For this problem we can use the standard Lorentz transformation:

t = t′γ , Where γ = 1√1− v2

c2

And we can use the Energy equation:

E = γm0c2

From the energy equation and solving for γ we find:

γ = ELabm0c2

= 10000498

Therefore, the answer to how long the particle exists in the lab frame is:

tLab = trestγ ≈ 20 ∗ 10−10[sec]

The velocity of the Kaon is:

v =√

1− 1γ2≈ 0.99876 c

Therefore the final answer is:

d = vt = 0.99876 ∗ 20 ∗ 10−10 ∗ (3x108) ≈ 0.599[meters]

(b) In the center of mass frame, the resulting velocity of each π0 is:

E = γm0c2 =⇒ v =

√1− 4

m2π

m2K≈ 0.84027 c

Using the velocity addition formula, we can find the velocity difference betweenthe lab frame and the frame of the pion. The two velocities are in oppositedirections.

s = u−v1−uv

c2≈ 0.9858 c⇒ γ ≈ 5.956

KE = mπc2(γ − 1) ≈ 675[MeV ]

(c) If the two pions have equal energy in the lab frame then the pions decayed in aplane perpendicular to the velocity of the original kaon. Therefore the decayangle is defined by the four-momentums.

(Pα,K)2 = (Pα,π1 + Pα,π2)2 ⇒ m2K = 2m2

p + 2(E2π − p1p2 cos θ)

Solving for θ

Problem ”problem13modern” from 2008Comps.

θ = arccos(m2π+

E2

4−m2K2

p1p2)

p1p2 = p2π = E2

4−m2

π

E is the original energy in the lab frame of the kaon.

thereforeθ ≈ 4.6degrees

5 Cosmic ray

Cosmic ray photons from space are bombarding your laboratory and smashingmassive objects to pieces. Your detectors indicate that two fragments, each ofmass m0, depart such a collision moving at a speed of 0.6c at angles of 60

relative to the photon’s original direction of motion.

(a) In terms of m0 and c, what is the energy of the cosmic ray photon?

(b) In terms of m0, what is the mass M of the particle being struck (assumedto be initially stationary)?

Solution

(a) The components of the fragment’s momenta in the direction perpendicular to theoriginal photon’s direction of motion will cancel, and the components along thatdirection will reinforce. Conservation of three momentum tells us that

pi = 2 pf cos(60o).

Using the relativistic definition of momentum, pf = mγ(vf )vf , where γ(vf ) =

1/√

1− v2f/c2 = 1.25,

pγ = pi = 2 pf cos(60o) =2m0v cos(60o)√

1− v2/c2.

Now since for a photon Eγ = pγc, then

Eγ = 2m0 · 0.6c2 cos(60o) · 1.25 =3

4m0c

2.

(b) Use conservation of relativistic energy, and the fact that the total relativisticenergy (including rest energy) of the final fragments is

Ef = m0γ(vf )c2 = 1.25m0c2

so thatEi = Eγ +Mc2 = 2Ef ,

and we have that

M =2Ef − Eγ

c2=

2.5m0c2 − 0.75m0c

2

c2= 1.75m0.

Problem ModQM-F01 from 2003-Proficiency-Spring.

6 Cosmic ray collision

A cosmic ray photon strikes a stationary particle of mass M, which fragmentsinto two pieces, each of mass m0. The two pieces depart the collision moving ata speed 3

5c at 60 to the photons original direction of motion. (c is the speed oflight in vacuum.)

(a) Determine the energy of the cosmic ray photon in terms of m0 and c.

(b) Determine M ,the mass of the particle being struck.

Solution

prcpr00

+

Mc2

000

=

√

(m0c2)2 + p2c2

p cos 60

p sin 60

0

+

√

(m0c2)2 + p2c2

p cos 60

−p sin 60

0

pr = p =

m0v√1−

(vc

)2 =v

c

m0c√1−

(vc

)2(a)

Er = pc =v

c

m0c2√

1−(vc

)2 =3

5

1√1−

(35

)2m0c2

Er =3

4m0c

2

(b)

Mc2 = 2

√(m0c2)2 +

(3

4

)2

m20c

4 − 3

4m0c

2

M =

(2 · 5

3− 3

4

)m0 =

7

4m0

7 Concepts

Provide a brief qualitative description for each item listed below.

(a) Heisenberg uncertainty principle.

(b) The fundamental forces.

(c) Michelson-Morley experiment.

Solution

(a) The Heisenberg uncertainty principle states that it is impossible to know simul-taneously the precise position and momentum, or time coordinate and energy,of a particle. The product of the uncertainties of the two measurements mustbe at least h/2. An example of the energy-time relation is the product of thelifetime of an unstable state and the uncertainty in the energy of the photon itemits (the decay line-width).

(b)In order of increasing strength these are: the gravitational force between anytwo massive objects, which falls off like 1/r2 and so has infinite range; theelectromagnetic force between any two charged particles, which has the sameforce law and infinite range; the weak force responsible for, for example, the betadecay of a neutron into a proton, and electron and an anti-electron-neutrino; andthe strong force which is responsible for binding quarks together into protons andneutrons and also for the nuclear force between nucleons. The weak force hasextremely short range because of the large mass of the force carriers (the W andZ bosons) and the strong force has effectively a short range because confinementof colored objects requires any final-state particles in an interaction to be colorneutral.

(c) This experiment was set up to determine how quickly the Earth moved throughthe aether which was thought to be the medium in which light propagated. Iflight moved with a fixed speed through the aether, the apparent speed of abeam of light through an apparatus fixed to the moving Earth would dependon whether it moved perpendicular or parallel to the direction of motion of theEarth. A beam of light was split by a half-silvered mirror and sent along twopaths at right angles to each other, reflected back by mirrors and then recom-bined. By counting the number of interference fringes moving by an observer asthe apparatus was rotated by 90, Michelson and Morley planned to measure thedifference in speed of the two beams. They saw no such motion of the fringes,which invalidated the idea of the aether.

Problem ModQM-F99-2 from 2003-Proficiency-Spring.

8 Photoelectric effect

When light of wavelength 520 nm is incident on the surface of a metal, electronsare ejected with a maximum speed of 1.78×105 m/s. What wavelength is neededto give a maximum speed of 4.81 × 105 m/s?

Solution

The electrons are bound to the metal with a work function φ, which is the minimumenergy required to remove an electron from the metal. Assuming an electron whichrequires only this minimum energy, the energy hν of the photon goes into releasingthe electron from the surface and any extra energy is kinetic energy of the electron,so that (note the electron velocity is non relativistic)

hν =hc

λ= φ+

1

2mev

2max.

For two different wavelengths we have

hc

λ1= φ+

1

2mev

21

hc

λ2= φ+

1

2mev

22 ,

where v1 and v2 are understood to be maximum velocities, and so we can eliminate φto obtain

hc

λ1− 1

2mev

21 =

hc

λ2− 1

2mev

22

hc

λ2=

hc

λ1+

1

2me(v

22 − v21)

or

λ2 =hc

hc/λ1 + 12me(v22 − v21)

= 1240 eV · nm/

1240 eV · nm

520 nm+

1

20.511× 106 eV

([4.81× 105]2 − [1.78× 105]2

)(3× 108)2

= 420 nm.

Problem ModQM-F99-3 from 2003-Proficiency-Spring.

9 Concepts 2

Provide a brief qualitative description for each item listed below.

(a) Cerenkov radiation.

(b) Planck’s constant.

(c) Rutherford scattering

Solution

(a) Cerenkov radiation occurs when a charged particle travels through a materialfaster than the speed of light c/n in that material. The radiation is emitted ina forward opening cone whose opening angle depends on the particle’s speed. Itis like the shock wave which occurs when a body moves faster than the speed ofsound in air.

(b) Planck’s constant h relates the energy of a photon to the frequency of the elec-tromagnetic radiation, E = hν. As such, it provides the link between the waveand particle nature of the photon (and other particles). Planck introduced h toexplain the failure of the Rayleigh-Jeans formula for black-body radiation, whichled to an infinity (the ultraviolet catastrophe) in the amount of energy emittedby a black-body when high energy photons were considered. His suggestion wasthat atoms emit energy only in discrete bundles whose energy was proportionalto the photon frequency.

(c) Rutherford scattered alpha particles (4He nuclei) off a gold foil. It was expectedthat there would be few backward scattering events because it was thought thatthe material in atoms was roughly evenly spread out. In fact the distributionof the scattering angle showed the existence of small massive centers of positivecharge in the gold foil. This led to the Rutherford scattering formula (for pointlike charged particles) and to the Bohr model of the atom.

Problem ModQM-F99-5 from 2003-Proficiency-Spring.

10 Muon

(a) How fast does a muon have to travel to have the same energy as a chargedpion at rest?

(b) What is the de Broglie wavelength of the muon?

(c) If you could measure the momentum of the muon with 10% accuracy, howwell could you measure its position along its direction of motion?

Solution

(a) The mass of the muon is mµ = 106 MeV/c2, and that of the pion is mπ = 140MeV/c2. The (total relativistic) energy of the moving muon is therefore

mµγc2 = mπc

2,

so thatγ =

mπ

mµ,

and so

1− v2

c2=

1

γ2=

(mµ

mπ

)2

v

c=

√1−

(mµ

mπ

)2

= 0.65.

(b) The de Broglie wavelength is given by

λ =h

p=hc

pc.

Now the momentum of a relativistic particle is p = mγv, so we have

λ =hc

pc=

hc

mγvc=hc√

1− (v/c)2

mc2(v/c)=

1240 MeV · fm√

1− (0.65)2

106 MeV · 0.65= 13.7 fm.

(c) According to the uncertainty principle we have that

∆p∆x ≥ h

2=

h

4π.

This means that

∆x ≥ h

4π(0.1p)=

h

4π(0.1h/λ)=

λ

0.4π= 10.9 fm.

Problem ModQM-F99-7 from 2003-Proficiency-Spring.

1 Muon decay

Radioactive decay proceeds according to the very simple law that the probabilityof decay for any given system is constant, that is

1

N

dN

dt= −λ. (1)

(a) Derive an equation for N(t), the number of particles remaining as a functionof time, in terms of N0 the number at time t = 0, the time t, and the decayconstant λ.

(b) In addition to λ, radioactive decay is often described in terms of the half lifet1/2 or the mean life τ at which times N(t) = 1/2N0 or N(t) = (1/e)N0,respectively. Derive expressions for t1/2 and τ in terms of λ as well as arelation between t1/2 and τ.

(c) The mean life of the muon is τ = 2.2 µs in its rest frame (proper time).Assume a group of muons is produced 8000 m above the surface of theearth and all travel directly downward at a speed of 0.995 c. (Use c =3×108 m/s). For this part, ignore special relativity. What is the flight timeto the surface of the earth classically and what is the surviving fraction(N(t)/N0) of muons?

(d) Now consider the proper treatment using special relativity and the Lorentztransformation from the point of view of an observer traveling with themuons. In this frame the mean lifetime is unchanged at τ = 2.2 µs. Whatdistance does this observer perceive as the distance to the earths surfaceand what is the travel time? Calculate the surviving fraction now.

(e) Lastly consider what an observer on earth would perceive in special rela-tivity. This observer would see that the muons must travel 8000m. Whatmean life would the earth based person observe and what would be thetravel time to the surface? Calculate the surviving fraction that the earthbased person would observe.

(f) Comment on the implications of parts (c) - (e) for a physicist on the grounddetecting muons. What does the principle of relativity and the indepen-dence of the frame of reference on physical outcomes say about what therelation between the answers for (d) and (e) should be?

Solution

(a) The decay is governed by the following differential equation

dN

N= −λdt, thus N(t) = N0 e

−λt. (2)

(b)

N(t1/2) = N0e−λt1/2 =

N0

2→ t1/2 =

ln(2)

λ. (3)

N(τ) = N0e−λτ =

N

e→ τ =

1

λ. (4)

λ =1

τ=

ln(2)

t1/2→ t1/2 = τ ln(2). (5)

(c)

t = L/v =8000m

0.995× 3× 108m/s= 26.8µs. (6)

N(t)/N0 = e−t/τ = e−26.8/2.2 = 5.12× 10−6

(d)

γ =1√

1− v2/c2= 10

L =L

γ= 8000m/10 = 800m

t =L

v=

800m

0.995× 3× 108m/s= 2.68µs.

N(t)/N0 = e−t/τ = e−2.68/2.2 = 0.30

(e) t = 26.8µs as in (c), but τ = γτ = 10 × 2.2µs = 22µs. N(t)/N0 = e−t/τ =e−26.8/22 = 0.30.

(f) Classical physics would predict a very tiny fraction of parts per million muonssurviving to the ground, whereas special relativity predicts 5 orders of magni-tude greater survival probability of 30% similar to what is observed. Also theobservers on the ground and traveling with the muon both predict the samesurvival probability.

One might wonder about the effect of general relativity due to Earth’s gravity. It isquite small. One can solve this problem exactly using the Schwarzschild metric

ds2 = c2dt20 =

(1− 2GM

rc2

)dt2 −

(1− 2GM

rc2

)−1

dr2. (7)

By setting M = 0 one immediately recovers t0 = t√

(1− v2/c2) which corresponds tothe above result. The correction from the gravity is very small

2GM

rc2=

2gr

c2∼ 10−9

for all distances ranging from the surface r = 6400km to 8km above the surface.

2 Thermodynamic cycle 2

Consider the following process done on an ideal monatomic gas. 1000 joulesof heat are added at constant pressure. The pressure is 1 atm. The systemexpands in the process and its temperature rises 10 K.

(a) Find the number of moles present.

(b) Find the increase in volume ∆V .

(c) Find the entropy change in terms of the initial temperature T0.

Solution

(a) According to first Law of Thermodynamics Q = ∆U + P∆V . For an idealgas PV = nRT and for a constant pressure process P∆V = nR∆T . For amonoatomic ideal gas U = (3/2)nRT , thus

Q = (3/2)nR∆T + nR∆T = (5/2)nR∆T

n = Q/((5/2)R∆T ) = 1000J/((5/2) ∗ 8.31J/(mole×K) ∗ 10K)) = 4.8moles

(b) P∆V = Q−∆U , and

∆U = 3/2nR∆T = (3/2)R∆TQ/((5/2)R∆T ) = 3Q/5

Thus P∆V = Q− (3/5)Q = (2/5)Q and

∆V = (2/5)Q/P = (2/5)1000J/(1.013× 105Pa) = 3.95× 10−3m3

(c) T∆S = Q = (5/2)nR∆T for the constant pressure process. ∆S = (5/2)nR∆T/T ,Integrating both sides

∆S = (5/2)nR lnT/T0 =Q

∆Tln

(T0 + 10K

T0

)= 100 ln

(T0 + 10K

T0

)

Problem 1 from Qualifying exam 2011-Spring.

3 4-momentum conservation

An energetic photon hits a stationary particle of rest mass M which breaksinto two equal mass particles of rest mass m0 each. The two particles leave thecollision with equal velocities 0.6c at equal angles of 60 degrees relative to thedirection of the incoming photon. Express all your answers below in terms ofm0 and c.

(a) What is the magnitude of the relativistic momentum of one of the outgoingparticles?

(b) Use relativistic momentum conservation to find the magnitude of the mo-mentum of the incoming photon.

(c) What is the energy of the incoming photon?

(d) What is the total (including rest mass) energy of one of the outgoingparticles?

(e) Use relativistic energy conservation to find the rest mass M of the struckparticle.

Solution

(a)

γ = 1/√

1− v2/c2 = 5/4, pf = m0γ0.6c =3

4m0c

(b) Momenta of outgoing particles perpendicular to photon direction must add to 0,momenta in direction of photon must add to equal its value:

pγ = 2pf cos(60) =3

4m0c

(c)

Eγ = pγc =3

4m0c

2

(d)

Ef = m0γc2 =

5

4m0c

2

(e)

Eγ +Mc2 = 2Ef , M = (2Ef − Eγ)/c2 =7

4m0.

Problem 1 from Qualifying exam 2012-Spring.

4 Pb nucleus

Using Einstein’s mass-energy relation, calculate the fraction of the mass of the208Pb (Z = 82) nucleus that can be attributed to the Coulomb interaction. Forthis estimate assume that the lead nucleus is a charged sphere of radius R =7fm = 7× 10−15m. The charge is uniformly distributed throughout the volumeof the sphere. Assume that the total nuclear mass is 1 amu=931MeV/c2 =1.66 × 10−27kg per each nucleon.

(a) What is the electric field, both outside and inside the sphere?

(b) What is the electrostatic potential at all points in space?

(c) What is the electrostatic energy of this uniformly charged sphere?

(d) Compute the 208Pb nuclear mass and the Coulomb energy from part (c),and find the fraction of the mass due to the Coulomb interaction.

Constants: e = 1.6×10−19C; k = 1/(4πε0) = 8.9×109Nm2/C2;α = 1/137; hc =197MeV · fm; c = 3 × 108m/s.

Solution

(a) Let Q be the total charge of the nucleus: Q = Ze. Using Gauss’ law we find

~E =1

4πε0

Q

r3~r, (r ≥ R)

~E =1

4πε0

Q

R3~r, (r < R)

(b) The electrostatic potential at any point p is given as an integral

φ(p) = −∫ p

∞

~E · d~r =

∫ ∞p

~E · d~r

This form assures that the potential is zero at infinity. We separate regions ofintegration 1-interior, 2-exterior

φ(p) =

∫ R

p

~E1 · d~r +

∫ ∞R

~E2 · d~r

This results in:

φ1 =1

8πε0

Q

R(3− r2

R2), (r ≤ R)

φ2 =1

4πε0

Q

r, (r > R)

(c) The energy can be found as

U =1

2

∫d3rφ(r)ρ(r) = 2π

∫ R

0

φ1(r)ρ(r)r2dr =3

5

1

4πε0

Q2

R=

3

5

1

4πε0

Z2e2

R

Problem 5 from Qualifying exam 2012-Spring.

(d) The Coulomb energy of the led nucleus using α = e2

4πε0hc

U =3

5αhc

Z2

R=

3

5

1

137197MeV fm

822

7fm= 829MeV

The mass of the lead nucleus is 208 times 1amu which means that the Coulombenergy is about 0.4% of the mass.

f =829MeV

208(931MeV/c2) · c2 = 0.43%

5 Proton in 1d well

A proton (mass 938 MeV/c2) is in a 1-dimensional square well 100nm inlength, with walls that are infinitely high. The center of the well is at x = 0.

(a) Is the proton relativistic?

(b) What is the expression for the proton’s allowed energy levels? Calculatenumerically the energy of the ground level.

(c) Calculate the expression of the expectation value of the position of theproton squeared

⟨x2

⟩in the ground level.

Solution

(a) Let’s consider the uncertainty principle,

∆p∆x ≥ h

2(8)

Thus,

∆p ≥ h

2∆x=

hc

2∆xc

Since a proton is confined in a 100nm 1-dimensional square well, ∆x = 100nm.Also, hc = 197.33MeV · fm. Therefore,

∆pc = ∆(mpvc)

= ∆

(938

MeV

c2vc

)= ∆

(938MeV

v

c

)≥ hc

2∆xc

=197.33MeV · fm

2× 100nm= 0.99eV

Therefore,v

c∼ 1.05× 10−9

This means that the proton’s speed iv much smaller than the speed of light. So,the proton is not relativistic.

(b) Since V (x) = 0 for −L2< x < L

2, the Shroedinger equation is

− h2

2m

∂2ψ

∂x2= Eψ.

The solutions for this equation is: will have the form:

ψ(x) = A sin(kx) +B cos(kx),

Problem MP17 from FSU Qualifying Exam Wiki.

where k =√

2mE/h2. Considering the boundary conditions, we have:

ψn(x) =

A sin

(kn(x− L

2

)), −L

2< x < L

2

0, others

where the normalization constant A is√

2L

and

kn =nπ

L=

√2mEn

h2 .

Therefore,

En =n2π2h2

2mpL2

The ground state energy will be for n = 1

E1 =h2π2

2mpL2

=h2c2π2

2mpc2L2

=(197.33MeV · fm)2π2

2(938MeV )(100nm)2

= 2.05× 10−8eV

(c) The ground state wave function is:

ψ1(x) =

√2

Lsin

(π

(x

L− 1

2

))Therefore,

⟨x2⟩ =

2

L

∫ L2

−L2

ψ∗1x2ψ1dx

=2

L

∫ L2

−L2

x2 sin2

(π

L

(x− L

2

))dx

=2

L

∫ L2

−L2

x2 cos2(πLx)dx

=2

L

∫ L2

−L2

x2 1 + cos(2 πLx)

2dx

=2

3L

(L

2

)3

+1

L

∫ L2

−L2

x2 cos(

2π

Lx)dx

Using the formula∫x2 cos axdx = 2x

a2cos ax+

(x2

a− 2

a3

)sin ax we can solve the

integral

⟨x2⟩ =

2

3L

(L

2

)3

+1

L

∫ L2

−L2

x2 cos(

2π

Lx)dx

=L2

12+

1

L

[L2x

2π2cos

2πx

L+

(Lx2

2π− L3

4π3

)sin

2πx

L

]L2

−L2

=L2

2

(1

6− 1

π2

).

6 Lepton in magnetic field

A µ-lepton moving at relativistic speed v is produced in a proton-protoncollision. The µ-lepton moves in a 1.0T magnetic field such that the anglebetween B and v is 85. As viewed along the direction of B, the projected pathof the µ-lepton is a circle of radius 10m.

(a) What is the momentum (in eV/c) and the Lorentz factor γ of the µ-lepton?

(b) In the rest frame, the µ-lepton has a mean lifetime τ = 2.197 × 10−6s.In the laboratory frame, what is its mean lifetime? (For this calculation,ignore the acceleration of the µ-lepton).

(c) Describe the trajectory (either qualitatively or quantitatively) of the µ-lepton’s path. How long will it take for the µ-lepton to travel one completeorbit in the plane perpendicular to the magnetic field? In this time, howfar will it move in the direction of the magnetic field?

Solution

(a) The magnetic force will work like a centripetal force, then we can match the bothforces as:

~Fm = ~Fc

⇒ q~v × ~B = γm0v2

rr

⇒ qvB sin 85 = γm0v2

r⇒ qBr sin 85 = γm0v

The last expression is precisely the momentum p = γm0v, where m0 is the restmass of the lepton, therefore:

p = γm0v

= qBr sin 85

= (e)(1T )(10m) sin 85

= (e)(

1Vs

m2

)(10m)

(2.998× 108m

s

c

)sin 85

=(2.998× 109) sin 85

eV

c

= 2.987× 109 eV

c

From the definition of the momentum, we can solve for the velocity v:

Problem MP12 from FSU Qualifying Exam Wiki.

p = γm0v

⇒ v√1− v2/c2

=p

m0

⇒ v =

(m2

0

p2+

1

c2

)−2

⇒ v =

((0.106× 109 eV

c2

)2c2(

2.987× 109 eVc2

)2 + 1

)−2

c

⇒ v = 0.997c (9)

Therefore,

γ =1√

1− v2

c2

= 12.92.

(b) We need to calculate the dilatation of the time for the laboratory frame:

τ ′ = γτ

= (12.92)(2.197× 10−6s

)= 28.38× 10−6s

(c) The trajectory will be like a “curl”. In the plane perpendicular to the magneticfield, it will be a circular movement, and in the parallel direction to the magneticfield, it will be a movement with constant velocity. The time for one loop willbe:

t =2πr

v

=2π (10m)

(0.997)(3× 108m

s

)= 0.23× 10−6s

The travel distance in this time in the direction parallel to the magnetic fieldwill be:

d = (v cos 85) t

= (0.997)(

3× 108m

s

)cos 85

(0.23× 10−6s

)= 5.94m

7 Bohr model

(a) Calculate the frequency of revolution and the orbit radius of the electronin the Bohr model of hydrogen for n = 100, 1000, and 10,000.

(b) Calculate the photon frequency for transitions from the n to n-1 statesfor the same values of n as in part (a) and compare with the revolutionfrequencies found in part (a).

(c) Explain how your results verify the correspondence principle.

Solution

(a) We’ll start by calculating the Bohr radius, which we can derive by Newton’ssecond law and the Bohr quantization rule for angular momentum:

F = mear = mev2

r

F =e2

4πε0r2

mev2

r=

e2

4πε0r→ mev

2 =e2

4πε0r

L = r × p = merv = nh→ v =nh

mer

mev2 = me

n2h2

m2er2

=n2h2

mer2

n2h2

mer2=

e2

4πε0r

rn =4πε0e2

n2h2

me

r100 = .5293µm

r1000 = 52.93µm

r10000 = 5290.µm

The radii for n = 10000 and n = 1000 would be easily visible to the naked eye.The radius for n = 100 would be easily visible with a weak microscope.

It is unclear what they’re looking for when asking for “frequency of revolution.”They could be asking for either ω (calculated here) or ν, which are related byω = 2πν.

ω =v

r=

nh

mer2

Using the radii found above we have:

ω100 = 4.13× 1010Hz

ω1000 = 4.13× 107Hz

ω10000 = 4.13× 104Hz

Problem Mod13 from FSU Qualifying Exam Wiki.

(b) The frequency of the photon will correspond to the change in energy between thetwo states. In case you don’t remember the equation for this energy it can bederived from the total energy of the electron:

E =1

2mev

2 − 1

4πε0

e2

r

E =1

2

e2

4πε0r− 1

4πε0

e2

r= −1

2

e2

4πε0r

r =4πε0e2

n2h2

me→ E = −me

2h2

(e2

4πε0

)21

n2

E = − Rn2

where R is the Rydberg constant: R = 13.6eV .So the energy of the photon is given by change in energy between states:

hν = R

(1

n2f

− 1

n2i

)

ν =R

h

(1

n2f

− 1

n2i

)=

Rc

2πhc

(1

n2f

− 1

n2i

)ν100 = 6.68× 109Hz

ν1000 = 6.59× 106Hz

ν10000 = 6.582× 103Hz

In order to compare these values to the values obtained in part (a) we needto turn “ordinary” frequency into angular frequency: ω = 2πν Below lists theangular frequency from part (a) and corresponding angular frequency for part(b):(a) ω100 = 4.13×1010Hz (a) ω1000 = 4.13×107Hz (a) ω10000 = 4.13×104Hz(b) ω100 = 4.20×1010Hz (b) ω1000 = 4.14×107Hz (b) ω10000 = 4.13×104Hz

(c) The correspondence principle (formulated by Bohr) states that quantum mechan-ical systems reproduce classical results in the limit of large quantum numbers.As you can see above the frequency of the emitted radiation in the classical limit(b) is equal to the frequency of rotation of the electron around the nucleus (b)where we used the quantum principle of quantization of angular momentum.These numbers agree better with larger quantum number n.

8 Uncertainty principle

Use the Heisenberg uncertainty principle to show that

(a) electrons are unlikely to be constituents of atomic nuclei

(b) their existence within atoms is reasonable.

Solution

The Heisenberg uncertainty principle states

∆x∆p >h

2

We can write the energy as:

E =p2

2m

The mass of the electron is:

m = 9.109× 10−31kg

The size of a nucleus and an atom are approximately:

xnuc ≈ 10−15m

xatom ≈ 10−10m

Solving the uncertainty relation for p and substituting these values for x:

pnuc ≈ 5.27× 10−20J · s/m

patom ≈ 5.27× 10−25J · s/mSubstituting these values into the energy equation:

Eatom = 1.53× 10−19J = 0.9eV

Enuc = 1.53× 10−9J = 9.5× 109eV = 9.5GeV

Nuclear binding energy is on the order of MeV and electron (atomic) binding energy

is on the order of eV. Thus an electron with energy on the order of GeV could not be

bound to the nucleus. However, and electron with energy on the order of eV could be

bound within an atom.

Problem Mod18 from FSU Qualifying Exam Wiki.

9 Magnetic moment of electron

Part I) Originally the electron was thought to be a tiny charged sphere (wenow treat it as a point object with no space extensions). Find the equatorialspeed assuming the electron is a uniform sphere of radius 3x10−6nm. Comparethe answer with the speed of light.Part II) A cylinder of radius R, with mass M uniformly distributed throughoutthe volume. The angular speed, ω is about the longitudinal axis.

(a) Obtain ~L.

(b) If a charge Q is distributed uniformly over the curved surface only, find the

magnetic moment ~µ of the rotating cylinder. Compare ~µ and ~L to deducethe g-factor.

Solution

Part IWe know that the magnitude of angular momentum, L, is given by L = rp = rmv.We also know that for electrons (which are spin 1/2) that the angular momentum isequal to h/2. Therefore L = mrv = h/2.Plug in the values for h, the mass of the electron (.511 MeV/c2), and the radius of theelectron and you find that velocity is:v = 2x1011m/s, which is of course faster than the speed of light!!Part II

(a) ~L =

∫ρ~rx~vd3r, where ρ is the density, ie ρ = M/V .

Write ~v as ~v = ~ωx~r = ωzx (rr + zz) = ωrφ

~L = ρ

∫(rr + zz) x

(ωrφ

)d3r = ρ

∫ [(rr + ωrφ

)+(ωrφ+ zz

)]d3r

~L = ρ

2π∫0

dφ

L/2∫−L/2

dz

R∫0

rdr[r2ωz − zωrr

]. Note that the second term goes to

zero when you do the z integration.

~L =M

V2πz

L/2∫−L/2

dz

R∫0

r3ωdr =MωR2z

2

This solution is correct since L = Iω if the angular momentum vector is parallelto the angular velocity vector and for a cylinder I = MR2/2.

(b) ~µ = 1/2

∫~rx ~Jd3r

~J = ρ~v; where again, ~v = ~ωx~r = ωzx (rr + zz) = ωrφ,

and, ρ =Q

2πRLδ(r −R)θ(Z ± L/2)

Plugging in these values and performing the integral as before yields:

Problem Mod11 from FSU Qualifying Exam Wiki.

~µ = QR2ωz/2

To solve for the g-factor use: ~µ =−gQ2M

~L and plug in the values for ~L and ~µ

found above. Surprisingly g = 2, which is approximately the g-factor for theelectron.

10 Short questions

The following questions are to be answered by a single number (order ofmagnitude) or by brief single sentence statements.

(a) What is the ionization potential of the hydrogen atom (in eV)?

(b) What is the electrostatic energy (in eV) of two electronic charges spaced1A apart?

(c) A capacitor has two cracks in the dielectric as shown. At which of the twocracks will breakdown occur first as the voltage is increased? Explain.

(d) Why is the sky blue and the sunset red?

(e) What is the range of wavelengths of visible light?

(f) What is the ratio of the magnetic moment of the electron and the proton?

(g) Are most substances (e.g. water, benzene) paramagnetic or diamagnetic?Explain.

Solution

(a) 13.6eV;

(b)e2

4πε0r=

(1.602× 10−19)2

4× 3.1415× 8.8542× 10−19 × 1× 10−10= 14.4eV ;

(c) B breaks down first. Suppose E is the electric field inside the capacitor, thenEA = E (Et continuous), EB = εE/1 = εE (Dn continuous). Thus EB > EAand B breaks down first.

(d) Rayleigh scattering σ ∝ k4: the relative percentage of short waves such as blueis heavier in the scattered lights, while in the forwarding lights, the relativepercentage of longer waves such as red is heavier.

(e) 3900A-7800A;

(f) µe/µn = mp/me ∼ 1837;

(g) Diamagnetic. Electrons gyrorate in the direction of externally applied field; sincethey carry negative charges, the induced magnetic moments are in the oppositedirection.

Problem 1 from UCLA 1988Fall.

1 Two level system

The D0 are particle and anti-particle. The wave function that describes aneutral D can be described as two component wave function.

|ψ〉 =

[D0

D0

]= D0

[10

]+ D0

[01

]where |D0〉 and |D0〉 are the particle and antiparticle components. The timeevolution of such a wave function can usefully be written

i∂

∂t|ψ〉 = H|ψ〉 = (M− i

2Γ)|ψ〉

with the Hamiltonian

H =

[M − i

2Γ M12 − i2Γ12

M∗12 − i2Γ∗12 M − i

2Γ

]M and Γ are the 2 × 2 mass and decay matrices; CPT invariance requiresM11 = M22 = M and Γ11 = Γ22 = Γ. Assumming CP invariance (which isapproximately true), M12 and Γ12 are real. In units where h = c = 1,

(a) What are the eigenstates of this Hamiltonian (label them D1 and D2)?(b) What are the lifetimes (λ1 and λ2) and the masses (m1 and m2) of the twostates? (Hint:Solve for the complex eigenvalues of the two states.)(c) Assuming the D is produced as a particle, |ψ(t = 0)〉, calculate the proba-bility as a function of time the particle is a D0 in the limit that ∆M , ∆Γ Γand ∆Mt, ∆Γt 1 where ∆M ≡ |m1 −m2|, ∆Γ ≡ |λ1 − λ2|.(d) At what time is |〈D0|ψ〉|2 maximum?

Solution

(a)

(M − i

2Γ− E)2 − (M12 −

i

2Γ12)2 = 0

Solving for E we have.

E± = M − i

2Γ± (M12 −

i

2Γ12)

Now we are going to find the eigenstates

H =

[M − i

2Γ M12 − i

2Γ12

M∗12 − i2Γ∗12 M − i

2Γ

] [ab

]=

(M − i

2Γ± (M12 −

i

2Γ12)

)[ab

]Choosing a = 1 we have

b = 1

Problem G3P from 2008Comps.

Then the eigenstate is

|D1〉 =1√2

[11

]For the other eigenvalue we have

|D2〉 =1√2

[1−1

](b) Looking at the eigenstates we have

λ1 =1

2(Γ + Γ12)

λ2 =1

2(Γ− Γ12)

and for the masses

m1 =1

2(M +M12)

m2 =1

2(M −M12)

(c) We can write |ψ〉 ad a linear convination of |D1〉 and |D2〉 as

|ψ〉 =1√2e−iE+t|D1〉+

1√2e−iE−t|D2〉[

D0

D0

]=

[12e−iE+t + 1

2e−iE−t

12e−iE+t − 1

2e−iE−t

][D0

D0

]=

[12e−i(M−

i2

Γ+M12− i2 Γ12)t + 12e−i(M−

i2

Γ−M12+ i2

Γ12)t

12e−i(M−

i2

Γ+M12− i2 Γ12)t − 12e−i(M−

i2

Γ−M12+ i2

Γ12)t

]So the probability is

P =1

4

[e−i(M+M12)t− 1

2(Γ+Γ12)t + e−i(M−iM12)t− 1

2(Γ−Γ12)t

] [ei(M+M12)t− 1

2(Γ+Γ12)t + ei(M−iM12)t− 1

2(Γ−Γ12)t

]After some algebra we get

P =1

4

(e−(Γ+Γ12)t + e−[Γ−i(m1−m2)]t + e−[Γ+i(m1−m2)]t + e−(Γ−Γ12)t

)Simplifying we have that the probability is given by

P =1

2e−Γt [cosh (Γ12t)− cos (∆Mt)]

(d) Taking the derivative of the probability wih respect to time we have

∂P

∂t=

Γ

2eΓt [cosh (Γ12t)− cos (∆Mt)] +

1

2eΓt [Γ12 sinh (Γ12t) + ∆M sin (∆Mt)]

To find the maximum we do the derivative equal to zero. And since Γ12t and ∆Mtare small we can taylor expand this to second order, so we get

−Γ

2(Γ12 + ∆M) t2 + (Γ12 + ∆M) t = 0

Since t = 0 is the minimum the maximum should be

t =2

Γ

2 Proton and antiproton

An antiproton with Kinetic Energy= 1.00GeV collides with a stationary pro-ton to form a final state which is a new particle, X0. In the following pleasegive numerical answers correct to at least two significant figures. Protons andantiprotons have rest-mass energy mpc

2 = 0.94GeV .(a) Calculate the rest-mass energy and velocity of the X0 particle. You mayexpress the velocity in units of c.(b) The X0 subsequently decays into two photons. One of the photons is emittedalong the direction in which the X0 was moving. Find its energy and the energyand direction of the other photon.(c) Another similarly produced X0 decays as shown below, with the photonsemitted at equal angles, β, to the direction in which the X0 was moving. Findthe energies of the photons and the angle β.

Solution

(a) Let T = 1.0GeV and c = 1 Using conservation of 4-momentum

(mp,~0) + (Ep, ~p) = (EX , ~px). (1)

Squaring both sides and using E2 − p2 = m2

m2p +m2

p + 2Epmp = m2x (2)

rearranging and using Ep = T +mp we get

mx =√

2mp(Ep +mp) (3)

mx =√

2mp(2mp + T ) (4)

Plugging in for mp and T .

mx =√

2 ∗ 0.94GeV (2 ∗ 0.94GeV + 1.0GeV ) = 2.3GeV (5)

And to determine the velocity we use

v =pxEx

=p

Ex=

p

T + 2mp=

√E2p −m2

p

T + 2mp=

√(T +mp)

2 −m2p

T + 2mp(6)

Plugging in for mp and T .

v =

√(1 + 0.94)2 − 0.942

1 + 2(0.94)

GeV

GeV= 0.59 (7)

(b) Again using conservation of 4-momentum

(Ex, ~px) = (pγ , ~pγ) + (p′γ , ~p′γ). (8)

Rearranging and using the fact that the angle between ~px and ~pγ is 0

(Ex, ~px)− (p′γ , ~p′γ) = (pγ , ~pγ). (9)

Problem G4P2 from 2008Comps.

Squaring and solving for pγ

m2x − 2Expγ + 2 ~px ~pγ = 0 (10)

m2x − 2Expγ + 2pxpγcos0

0 = 0 (11)

m2x − 2pγ(Ex − px) = 0 (12)

pγ =m2x

2(Ex − px)=

2mp(2mp + T )

2(T + 2mp − p)=

mp(2mp + T )

T + 2mp −√

(T +mp)2 −m2p

(13)

pγ =mp

1−√

TT+2mp

(14)

and plugging in for mp and T we get.

pγ =0.94GeV

1−√

11+2(0.94)

= 2.3GeV = Eγ (15)

For the proton going in the opposite direction

E′γ = p′γ = Ex − pγ = T + 2m− p− Eγ = 1GeV + 2(0.49GeV )− 2.3GeV = 0.59GeV(16)

(c) In this case the decay is symmetric so

pγ = p′γ =Ex2

= Eγ = E′γ =T

2+mp = 0.5GeV + 0.94GeV = 1.4GeV (17)

Again using 4-momentum

(Ex, ~px)− (p′γ , ~p′γ) = (pγ , ~pγ). (18)

Squaring then plugging in for pγ gives

m2x − 2Expγ + 2pxpγcosβ = 0 (19)

cosβ =E2x −m2

x

pxEx=pxEx

= v (20)

β = cos−1(v) = cos−1(0.59) = 540 (21)

3 Paint gun

Astronauts Sally and Bob decide to settle one of their disagreements in a ”paintduel”. They fly their space shuttles on a collision course toward each other whilekeeping their shuttles at constant speed. Each shuttle is equipped with paintguns which are capable of emitting spray paint at high speed. Whoever gets themost paint on their shuttle loses. A worker from from space station DS3 is thereto observe and referee the duel. Note: Express times in units of s (seconds),velocities in units of c (speed of light), and distance in units of light seconds= cs.In the frame of the space station the following events take place:event 1.) Sally fires her paint gun at x = 9cs and t = −2s.event 2.) Bob fires his paint gun at x = 0cs and t = 0s.event 3.) Bob’s ship is hit by paint at x = 4cs and t = 5s.event 4.) Sally’s ship is hit by paint at x = 5cs and t = 6s.NOTE: For this problem we let c = 1.The 4-vectors for the events are as follows:

S1 = (−2, 9, 0, 0)s (22)

S2 = (0, 0, 0, 0)s (23)

S3 = (5, 4, 0, 0)s (24)

S4 = (6, 5, 0, 0)s (25)

(a) Show that Sally’s velocity in the rest frame of the space station is vS =−1/2c while Bob’s velocity is vB = +4/5c.

(b) Find the intervals ∆x′ and ∆t′ between events 1 and 2 in Sally’s rest frame.

(c) Determine if it is possible for events 1 and 2 to be causally connected.

(d) Determine the velocity of the paint from Bob’s paint gun in Bob’s restframe.

(e) Bob is trying to evade the rules of the paint duel, and sends a massageon his communicator to a friend’s shuttle hidden behind Sally’s shuttle.Bob’s communicator is using a frequency of f = 1010Hz. Sally’s receiveris able to receive communications in the frequency range of 3 × 109 to3× 1010Hz. Determine whether Sally is able to receive Bob’s message.

Solution

(a) Using the two 4-vectors for Sally i.e. S1 and S4 we can determine Sally’s velocity,Vs, in the frame of DS3.

Vs =∆x

∆t=

(9− 5)s

(−2− 6)s=

4

−8= −1

2(26)

Problem G4P7 from 2008Comps.

Using S2 and S3 we can determine Vb.

Vb =∆x

∆t=

(0− 4)s

(0− 5)s=

4

5(27)

(b) Using Lorentz transformations for space and time we get

x′ = γ(x− βt)⇒ ∆x′ = γ(∆x− β∆t)

t′ = γ(t− βx)⇒ ∆t′ = γ(∆t− β∆x)

Plugging in for S1 and S2 and using from part a.) Vs = −1/2 we get

∆x′ =1s√1− 1

4

(9 +1

2(−2)) =

2√3

8s =16√

3s

∆t′ =1s√1− 1

4

(−2 +1

2(9)) =

5√3s

(c) The easiest way to determine if two events are causally connected is to square theinterval ∆S = (∆t′,∆~x′). If the interval is timelike, ∆t′ > ∆x′; the two eventscan be causally connected. If the interval is spacelike, ∆x′ > ∆t′; the pair ofevents cannot be causally related. This is because events that have a spacelikeinterval happen too far away in space and near in time so that an inertial framewould have to travel faster than light in order to transport a clock betweenevents. Our interval is

(∆S)2 = (∆t′)2 − (∆x′)2 = (5√3s)2 − (

16√3s)2 (28)

Seeing ∆x′ > ∆t′; we have a spacelike interval. Therefore, events 1 and 2 cannotbe causally connected.

(d) We know Bob’s speed relative to DS3, and we can easily determine Bob’s paint’sspeed using S1 = (0, 0, 0, 0)s (where the paint started) and S4 = (6, 5, 0, 0)s(where the paint hit Sally’s shuttle).

Vp =∆x

∆t=

(0− 5)s

(0− 6)s=

5

6(29)

And to determine the paint’s speed, (v′), in Bob’s frame we just use the defini-tions of velocity and Lorentz transformations as follows:

V ′ =dx′

dt′=γ(dx− βdt)γ(dt− βd)

=dxdt− β

1− β dxdt

=Vp − β1− βVp

(30)

We are using c = 1 so β = Vb, so plugging in we get:

V ′ =56− 4

5

1− 56

45

=1

10(31)

(e) The frequency Sally detects is

f =1

λ=

1

∆t− v∆t, (32)

(where v is the relative velocity between Sally and Bob) and rewriting the properfrequency we get

f0 =1

λ0=

1

τ=

γ

∆t(33)

Dividing equation (11) by equation (12):

f

f0=

∆t

γ(∆t− v∆t)=

√1− v2

1− v =

√1 + v

1− v (34)

so now all we need to do if find the relative velocity between Sally and Bob, andwe do this in the same way as in part d.)

v =Vb − Vs1− VsVb

=45

+ 12

1 + 45

12

=13

14(35)

Plugging this into equation 13 and solving for f :

f =

√27/14

1/14f0 =

√27f0 = 3

√3f0 (36)

f0 = fb = 1010Hz so f = 3√

3× 1010Hz. This means Sally cannot tell that Bobis trying to cheat, because the frequency is just a bit too high for her receiver!!

4 Neutrino beams

(µ-Neutrino Beams) A beam of high energy µ-neutrinos is made at FermiLabby forming a monoenergetic 2 GeV π+ beam, allowing the pions to decay bythe process π+ → µ+ +νµ, and then deflecting the µ+ particles out of the beamwith a strong magnet.

(a) Find the energy of neutrinos produced by this process in the rest frame ofthe pions.

(b) Find the energy of a neutrino produced at an angle of 0 with respect tothe incident pion direction in the laboratory frame.

(c) Find the laboratory-frame emission angle θ (measured with respect to theincident pion direction) at which the neutrino energy falls to half theenergy of neutrinos at θ = 0, as calculated in part ii above.

Solution

(a) Four-momentum conservation in the CM frame

(mπ 0) = (Ecmµ ~p) + (Ecmν − ~p) (37)

(mπ 0)− (Ecmν − ~p) = (Ecmµ ~p)

m2π − 2mπE

cmν = m2

µ

Ecmν =m2π −m2

µ

2mπ=

(134.98 MeV )2 − (105.66 MeV )2

2 · 134.98 MeV= 26.14 MeV.(38)

(b) Four-momentum conservation in the lab frame

(Eπ ~pπ) = (Eµ ~pµ) + (Eν ~pν) (39)

(Eπ ~pπ)− (Eν ~pν) = (Eµ ~pµ)

m2π − 2EπEν + 2pπEν cos (θ) = m2

µ (40)

Eν =m2π −m2

µ

2(Eπ − pπ cos (0))=

m2π −m2

µ

2(Eπ −√E2π −m2

π)

Eν =(134.98 MeV )2 − (105.66 MeV )2

2(2000 MeV −√

(2000 MeV )2 − (134.98 MeV )2)= 774 MeV.

(c) Solving equation 40 for θ and using Eν2

for the new value of the neutrino energywe get

θ = cos−1

(m2µ −m2

π + EπEν

Eν√E2π −m2

π

)θ = cos−1

((105.66 MeV )2 − (134.98 MeV )2 + 2000 MeV · 774 MeV

774 MeV√

(2000 MeV )2 − (134.98 MeV )2

)= 3.87.

Problem G4P18 from 2008Comps.

5 Star around cluster

How does the orbital velocity ~v of a bright star that is in a circular orbitaround a cluster of galaxies depend on the orbit radius ~r and the system massM , under the following assumptions?

(a) Under the assumption that the total attractive mass MV is in the form of

visible stars that are distributed symmetrically within a radius ~RV < ~r.

(b) Under the assumption that the mass of visible stars is negligible and thatthe orbit occurs within a uniform sphere of dark matter of mass MD witha radius RD > ~r.

(c) It is observed that for a number of bright stars orbiting at various distancesaround a galactic cluster, v(r) ∼ r and that v2r >> GMV . Discuss theseresults in the context of evidence for dark matter.

Solution

(a) To do this part all we have to do is balance the force due to gravity and thecentripetal force due to the stars orbit.

GMvM

r2=Mv2

r(41)

and solve for v

v =

√GMv

r(42)

(b) For this part we assume that the only mass the affects the orbit is that which iswithin the orbit radius, and we know it is a uniform sphere so:

ρ(r) = ρ(RD). (43)

Now just using the the fact that ρ = M/V , and letting the inner mass be MDr

we get:MDr

Vr=MD

VD⇒MDr =

MDr3

R3D

(44)

Plugging this into our forces equation gives us

GMDrM

r2=GMDMr3

r2R3D

=Mv2

r(45)

and solve for v

v = r

√GMD

R3D

(46)

(c) For this part we add the forces from the visible stars and the dark matter like so

GMvM

r2+GMDMr3

r2R3D

=Mv2

r(47)

Problem G5P12 from 2008Comps.

rearranging and multiplying by r

v2r = GMv +r3

R3D

GMD (48)

So if Mv is negligible compared to MD (and the orbit is not close to the centeri.e. r RD is not true) then yes, v2r >> GMV . v(r) ∼ r can be seen fromequation (6). So if this is how observed stars act then it does indeed point todark matter.

6 Superluminal objects

Studies of the quasar 3C345 using long-baseline radio telescopes have shownthat the object emitted a radio-bright ”fireball” that, in observations madebetween 1969 and 1976 separated from the main body of the quasar and movedan apparent distance of 56 light years during the observations. Taken at facevalue, the hot object would be traveling at 8 times the speed of light. Assumethat the quasar is a distance D from the Earth and that the hot object is movingaway from the quasar in the general direction of the Earth at a relativisticvelocity βc, with its direction of travel making a small angle θ with the line ofsight between the quasar and the Earth.

(a) Calculate the actual separation distance ~R between the quasar and fireballand the apparent separation distance ∆x from the quasar, as seen by anobserver on Earth, as the hot object moves away from the quasar in thetime interval t. Here ∆x is the component of ~R perpendicular to the lineof sight, and t, ~R and ∆x are in the reference frame of an observer onEarth.

(b) Calculate the transit time T1 to the Earth of radio waves emitted whenthe object is ejected from the quasar, the transit time T2 of radio wavesemitted by the hot object a time t later, and the time difference ∆Tbetween the arrival at Earth of the waves from the two events.

(c) Assuming that θ is a small angle, use appropriate approximations to esti-mate the apparent velocity ~v of the object. Explain how it can exceed thevelocity of light.

Solution

(a) Let c = 1. Then R = βt, and ∆x = βt sin θ.

(b) If the distance is D then the time it takes the radio waves to get to earth isT1 = D/c = D, and for T2 we need to take into account how for the fireball hasmoved towards us in the time, t that has elapsed. So, T2 = D − βt cos θ, then∆T = D −D + betat cos θ = betat cos θ.

(c) The apparent velocity can be found by comparing distances. The apparent dis-tance is the apparent velocity times the time it takes the light to get there,Vateye, and this is equal to the real time times the real velocity.

Vateye = βt (49)

or plugging in for t = D and teye = D − βt cos θ.

Va(D − βD cos θ) = βD (50)

solving for Va

Va =β

1− β cos θ(51)

Problem G5P12-2 from 2008Comps.

Which for the small angle approximation is just

Va =β

1− β . (52)

So if θ is small and β is more than 12

then the apparent velocity is in fact largerthen the speed of light.

7 Two protons

Two protons approach each other head on at speeds of 0.5c relative to thereference frame S.

(a) Find the speed of one proton in the rest frame S′ of the other. Recall thatthe relativistic addition rule for velocities in, say, the x-direction is givenby:

u′x =ux − v

1− uxv/c2(53)

where v is the relative velocities of S and S′.

(b) What is the kinetic energy of the two protons in the frame S.

(c) What is the kinetic energy of the two protons in the frame S′.

(d) What is the momentum of the moving proton in the frame S′. The massof the proton is 938.28 MeV.

Solution

(a)

ux =(−0.5c)− (0.5c)

1 + (0.5c)(0.5c)

c2

= 0.8c (54)

(b) We know E = KE +m and that E = γm, so KE = γm−m = (γ − 1)m. Alsolet c = 1.

KE = (γ−1)m =

(1√

1− (0.5)2− 1

)m = (0.1547)(938.34MeV ) = 145.16MeV

(55)

(c) If we are in the S′ frame then of course the KE of one of the protons is zero, andto find the KE of the other we just use the same formula as we used in part b)and the relative velocity found in part a).

KE = (γ−1)m =

(1√

1− (0.8)2− 1

)m = (0.667)(938.34MeV ) = 625.87MeV

(56)

(d)p = γmv = (1.667)(983.34MeV )(0.8) = 1251.12MeV (57)

Problem G5P17 from 2008Comps.

8 K mesons

K+ mesons can be photoproduced by the reaction

γ + p→ K+ + Λ

(a) Determine the threshold (minimum) photon energy, as measured in thelaboratory (the rest-frame of the proton),for the above reaction to occur.

(b) Determine if it is possible for either

(i) the K+ or

(ii) the Λ to be at rest in the lab frame, and determine for whatphoton energy this could happen.

Solution

(a) Using the invariant mass of the initial system in the lab frame we have

M2 = m2p + 2Eγmp (58)

After the colission we get the invariant mass in the center of mass. Since we wantthe threshold energy the particles are at rest in the center of mass, therefore

M2 = (mK+ +mΛ)2 (59)

from equation 1 and 2 we get

Eγ =(mK+ +mλ)2 −m2

p

2mp(60)

Pluging the values in this equations we get that

Eγ = 911MeV

(b)

(i) First we write the four vectors for the reaction in the lab frame:

(Eγ , Eγ) + (mp, 0) = (mK+ , 0) + (EΛ, pΛ) (61)

To pretty up the math, we move the K+ term to the other side before squaring.

[(Eγ , Eγ) + (mp, 0)− (mK+ , 0)]2 = (EΛ, pΛ)2 (62)

mp2 +mK+

2 + 2Eγmp − 2EγmK+ − 2mpmk = m2Λ (63)

And solving for Eγ :

Eγ =m2

Λ

2(mp −mK+)− mp −mK+

2(64)

Problem modern1 from 2008Comps.

Plugging in the rest-mass energies (in MeVc2

), We get

Eγ = 1178.05MeV

This energy is a positive number, and above the threshold energy. Therefore,producing a kaon at rest is possible.

(ii) Part (ii) follows the same as in part (i) except that our mK+ and ourmΛ are interchanged. Therefore we can immediately write down the solutionfor Eγ .

Eγ =m2K+

2(mp −mΛ)− mp −mΛ

2(65)

Which gives an energy ofEγ = −600.87MeV

This energy is negative, hence it is not possible to produce a stationary lambdaparticle in the rest frame of the proton.

9 Muons

A relativistic beam of monochromatic muons produced in the upper atmo-sphere is incident vertically on the earth’s surface at velocity v ≈ c, where c isthe speed of light.

(a) Find the ratio of muon number reaching the ground to the number at aheight H above sea level. Assume that the beam moves at a constantvelocity and that the attenuation of the beam is due entirely to the spon-taneous decay of the muons. Express your answer in terms of the energyEµ of a muon as measured by an observer at rest on the ground, the muonrest- mass mµ, and the muon rest-frame lifetime(i.e., decay time) τ0.

(b) Explain in one sentence why the ratio must be the same for an observercomoving with the beam. Re-express your answer to (1) entirely in termsof parameters measured by a comoving observer.

Solution

(a) The decay of muons dN at any time is directly proportional to the number ofmuons present at that moment N and to the differential time dt (where weconsider N to be constant over the interval dt). Thus,

dN ∝ −Ndt

dN is negative as dt increases, requiring a negative sign on the right hand side.

The proportionality constant is the probabilty for a muon to decay per unit timeinterval, hence it is 1/τ , where τ is the mean lifetime of the particle.

dN = − 1

τNdt

Doing the usual integration, we get

N = N0exp(−t/τ)

where N0 is the initial number of muons.

For a person on the earth, the rest lifetime of the muon τ0 is time dilated by afactor of γ. Hence,

τ = γτ0

The person on the earth measures the muon’s velocity vE as the ratio of thedistance travelled H to the time taken to travel the same, tE , hence

vE =H

tE

Hence, we the ratio as

N

N0= exp(−H/vEγτ0) = exp(−Hmµc

2/vEτ0Eµ)

where mµ is the rest-mass of the muon and Eµ is the energy of the muonmeasured by a person on the earth.

Problem ModernPhysics1 from 2008Comps.

(b) How can the ratio depend on the frame of the observer? The time intervals inthe frame of the earth and that of the muon are related by-

γtµ = tE

Hence,

vE =H

tE=

H

γtµ=vµγ

where vµ is the self-velocity measured by a person co-moving with the beam.We substitute for vE in terms of vµ in the ratio derived in Part(1). Hence,

N

N0= exp(−Hmµc

2/vEτ0Eµ) = exp(−H/vµτ0) = exp(−tµ/τ0)

where tµ is the time interval taken by the muons to fall through a height H asmeasured by a person co-moving with the muons.

10 Doppler effect

The radar speed detector operates on a frequency of 109 Hz. What is the beatfrequency between the transmitted signal and the one received after reflectionfrom a car moving at 30m/sec?

Solution

Suppose the car is moving towards the radar with velocity v. Let the radar frequencybe ν0 and the frequency of the signal as received by the car be ν1. The situation isthe same as if the car were stationary and the radar moved toward it with velocity v.Hence the relativistic Doppler effect gives

ν1 = ν0

√1 + v/c

1− v/c ' ν0

(1 +

v

c

)correct to the first power of v/c. Now the car acts like a source of frequency ν1, sothe frequency of the reflected signal as received by the radar (also correct to the firstpower of v/c) is

ν2 = ν1

√1 + v/c

1− v/c ' ν1

(1 +

v

c

)' ν0

(1 +

v

c

)2

' ν0

(1 +

2v

c

)Thus the beat frequency is

ν2 − ν0 = ν0 ·2v

c= 109 × 2× 30

3× 108= 200 Hz.

1 Short questions 2

(a) Estimate the collision frequency per molecule (in sec−1) for the gas in thisroom.

(b) Demonstrate the validity of the dipole approximation for typical atomictransitions; do this by computing the characteristic difference in the pho-ton phase φ over atomic dimensions.

(c) Photons and neutrinos (energy 10 MeV) are emitted simultaneously froma supernova at a distance of 1.5×1025cm. If the neutrinos arrive here 100sec after the photons, what is their mass?

(d) Neutral atoms interact by electrical forces. Explain qualitatively how thishappens and derive the form of the interaction at large separation.

Solution

(a) The mean free path of the molecules is

λ ∼ 1√2πd2n

where d is the diameter of the molecule and n is the number density of the gas.The mean speed is

v =

√8kT

πm,

so the frequency is

f =v

λ=

√8kT

πm

√2πd2

p

kT=√

16πpd2√mkT

.

Since p ∼ 106Pa, d ∼ 1A = 10−10m, k = 1.38 × 10−23J/K, T ∼300K, m ∼28×10−3Kg

6×1023,

f ≈ 5× 108/sec.

(b) Incoming EM wave (photons):

E = E0ei(k·r−ωt)

Since the dimension of atoms is r ∼ 1A, and λ ∼ 103A− 104A, then

k · r =2π

λr ∼ 10−3 1,

eik·r ∼ 1

which is dipole approximation.

Problem 6 from UCLA 1987Fall.

(c) The distance isD = ct1 = vt1 + v∆t,

so

∆v = c− v =c2∆t

D= 2× 10−13c.

The momentum of the neutrino is

p = γmv =E

c,

then

m =E

γvc=E

vc

√1− v2

c2≈ E

c2v

√2c∆v = 6.3eV/c2.

(d) The dipole field E ∼ 1r3

, and the induced dipole p ∼ 1r3

, thus the interaction

U ∼ p · E ∼ 1

r6.

2 Numerical values 2

Give approximate numerical values for the following (be sure to give units):

(a) Escape velocity from the earth

(b) Ground state energy of muonium (µ+, µ− atom)

(c)∑ (−1)n−1

n

(d) Spin of the deuteron

(e) Lifetime of the neutron

(f) Specific heat of Cu at room temperature (assumed greater than Debyetemperature)

(g) The orbital period of a planet (Saturn) which is 3 times as far from thesun as the earth is.

Solution

(a) √GM

Re∼ 7.9km/s

(b)

ε ∼ −µe4

2h2

where

µ =mµ+mµ−

mµ+ +mµ−≈ mµ

2∼ 105MeV/c2

2so

ε ∼ 13.6eV × 105/2

0.51≈ 1428eV.

(c)∞∑n=1

(−1)n−1

n= ln 2

(d) Spin of deuteron is S = 1.

(e) Life time of neutron is ∼ 1000s.

(f) ∼ 3R ≈ 6cal/(mol*K)

(g) Since R3/T 2 =const, so the period of Saturn is

1year × (3

1)3/2 =

√27year ∼ 5.2years.

Problem 2 from UCLA 1989Spring.

3 Neutron density flow

The diffusion equation∂U

∂t= α∇2U + βU

provides a crude model to describe the flow of neutron density U(t, x, y, z) in ablock of uranium. Here, α is the diffusion constant, and β governs the rate atwhich free neutrons are created through collision.

(a) Use separation of variables to obtain a solution in a cube of size D: 0 <x, y, z < D, given an arbitrary distribution U(0, x, y, z) = Φ(x, y, z) andthe boundary condition that U ≡ 0 on the surface of the cube for all t ≥ 0.

(b) Use the solution of part (a) to determine the critical size, Dc, of such acube such that U will grow exponentially with time if D > Dc.

Solution

(a) Let U = X(x)Y (y)Z(z)T (t), then

∂T

∂t/T = α(

X ′′

X+Y ′′

Y+Z′′

Z) + β = λ2

⇒ T ∝ eλ2t

alsoX ′′

X+Y ′′

Y+Z′′

Z=λ2 − βα

Let Z ∝ sin kxx, Y ∝ sin kyy, Z ∝ sin kzz,

⇒ k2x + k2y + k2z = −λ2 − βα

And U(t, x, y, z) = 0 at boundary,

⇒

kx = lπ

D

ky = mπD

kz = nπD

(1)

hence

(l2 +m2 + n2)π2

D2=β − λ2

αor

λ2 = β − α π2

D2(l2 +m2 + n2)

⇒ U(t, x, y, z) =∑l,m,n

Almn sinlπx

Dsin

mπy

Dsin

nπz

De[β−απ

2

D2 (l2+m2+n2)]t

with U(0, x, y, z) = Φ(x, y, z),

Almn =23

D3

∫ ∫ ∫Φ(x, y, z) sin

lπx

Dsin

mπy

Dsin

nπz

Ddxdydz

=8

D3

∫ D

0

dx

∫ D

0

dy

∫ D

0

dzΦ(x, y, z) sinlπx

Dsin

mπy

Dsin

nπz

D

Problem 19 from UCLA 1990Spring.

l,m, n = 1, 2, 3, ...

(b) For exponentially grow,

β − απ2

D2(l2 +m2 + n2) > 0

⇒ β >3απ2

D2

⇒ D2 >3π2α

β= D2

c

⇒ Dc = π

√3α

β.

4 Model for a star

Consider as a hypothetical model for a star an ideal gas sphere in hydrostaticequilibrium at a uniform temperature T0. Assume spherical symmetry.

(a) Find an equation for the pressure derivative dP/dr at radius r, in termsof the gravitational constant G, the density ρ(r) at radius r and the massinterior to r, M(r).

(b) Assume the star undergoes a uniform contraction: the distance betweenany two mass elements changes by the same fraction y. Find the changein the star’s temperature necessary to maintain hydrostatic equilibrium.You may neglect the effect of radiation pressure.

(c) Assuming the surface of the star emits radiation as a black body, and basedon your answer in (b): How will the color of this change as it contracts?How will the luminosity L change? (L =energy radiated per unit time).

Solution

(a) Equation for equilibrium

(P (r − dr)− P (r))4πr2 = GdmM(r)

r2

dm = ρdV = ρ4πr2dr

⇒ −dPdr

=GρM(r)

r2

(b) Change scale of the length r → r/y = r′, (y > 1), so ρ→ ρy3 = ρ′, M →M ,

−dP′

dr′= Gρ′M(r′)/r′2

−dP ′ = Gρy3M

r2/y2dr

y=GρM

r2dry4

⇒ dP ′ = y4dP, P ′ = y4P

Since P = nkT for ideal gas, P ′ = n′kT ′, n′ = y3n,

⇒ T ′ = yT ⇒ T ′ = yT0 ⇒ ∆T = (y − 1)T0

(c) When the temperature of the star changes, the radiation spectrum shifts. Nowy > 1, the star contracts, the temperature increases, from the formula

hνmax = 2.82kT,

we know νmax will increase, so the color of this star changes to higher frequencylight like blue.

L = σT 44πR2 = 4πσR2T 4

L′

L=T ′4R′2

T 4R2=y4

y2= y2,

the luminosity increases.

Problem 4 from UCLA 1990Fall.

5 Differential equation

(a) J(z) is a solution of the differential equation zJ ′′ + J ′ = zJ(z) = 0 withJ(0) = 1. Give the first two non-zero terms in the series expansion of J(z)about z = 0.

(b) Find the first two terms in the asymptotic expansion of

F (t) =

∫ π2

0

expt cos θθ2J(θ)dθ

as t→ +∞.

Solution

(a)zJ ′′ + J ′ + zJ(z) = 0

J(0) = 1

LetJ(z) = 1 + az + bz2 + cz3 + dz4 + ...

Put this into the equation, then

12dz3 + 6cz2 + 2bz + a+ 2bz + 3cz2 + 4dz3 + z + az2 + bz3 + ... = 0

⇒ a = 0, b = −1

4, c = 0, d =

1

64

⇒ J(z) = 1− 1

4z2 +

1

64z4 + ...

(b)

F (t) =

∫ π2

0

expt cos θθ2J(θ)dθ

when t→∞,

F (t) ≈∫ π

2

0

expt(1− θ2

2)θ2(1− 1

4θ2)dθ

F (t) = et∫ π

2

0

exp(− tθ2

2)θ2(1− 1

4θ2)dθ

The first term is

F (t)1 =et

(√t/2)3

∫ π2

√t2

0

exp(−θ′2)θ′2dθ′

=et

(√t/2)3

∫ ∞0

exp(−x2)x2dx

=et

t3/223/2(

1

2Γ(

3

2))

=

√π

2

et

t3/2

Problem 9 from UCLA 1990Fall.

The second term is

F (t)2 = −et∫ π

2

0

exp(− tθ2

2)θ2

1

4θ2dθ

= − et

4(√t/2)5

∫ π2

√t2

0

exp(−θ′2)θ′4dθ′

= − et

4(√t/2)5

∫ ∞0

exp(−x2)x4dx

= − et

4t5/225/2(

1

2Γ(

5

2))

= −3√

2π

8

et

t5/2

So the first two terms are

F (t) =

√π

2

et

t3/2− 3√

2π

8

et

t5/2

6 Asymptotic expansion

Consider the integral expression

J(a) =4a√π

∫ ∞0

dxxe−x

2

1− e−a/x

(a) What is the limiting asymptotic form of J(a) as a→ 0? Keep at least twoterms in the expansion.

(b) What is the limiting asymptotic form of J(a) as a → ∞? Keep at leasttwo terms in the expansion.

Solution

(a) We use a ∼ 0,

(1− e−a/x)−1 = (a

x− a2

2x2+

a3

6x3+ ...)−1

=x

a[1− a

2x+

a2

6x2+ ...]−1

=x

a[1 +

a

2x+ ...]

We have then

J(a) ∼ 4√π

∫ ∞0

dx(1 +a

2x)x2e−x

2

∼ 4√π

[

∫ ∞0

x2e−x2

dx+a

2

∫ ∞0

xe−x2

dx]

∼ 4√π

[

√π

4+a

2

1

2]

∼ 1 +a√π

(b)

a→∞, e−a/x 1

(1− e−a/x)−1 ∼ 1 + e−a/x

Hence

J(a) ∼ 4a√π

[

∫ ∞0

xe−x2

dx+

∫ ∞0

dxxe−(x2+a/x)]

We use saddle point in the second integral,

f(x) = x2 +a

x

f ′(x) = 2x− a

x2⇒ x0 = (

a

2)1/3

f ′′(x) = 2 +2a

x3

Problem 8 from UCLA 1992Fall.

f(x0) = (a

2)2/3 + 21/3a2/3 =

3

41/3a2/3

f ′′(x0) = 6

We have then∫ ∞0

dxxe−(x2+a/x) ∼∫ ∞0

dxx0e−f(x0)e−f

′′(x0)(x−x0)2

2

≈ x0e−f(x0)(2π

f ′′(x0))1/2

So

J(a) ∼ 4a√π

[1

2+ (

a

2)1/3(

π

3)1/2 exp(− 3

41/3a2/3)]

7 Model of the sun

In the following problem order of magnitude estimates will suffice. Model thesun as a uniform constant density (ρ = 1.4 g cm−3) sphere of ionized hydrogenwith radius R ≈ 7× 1010cm. For the purpose of estimating the internal energyof the sun, assume a characteristic temperature of T ≈ 4.5× 106K. (Note thatthis is not the surface temperature of the sun.) Further assume that localthermodynamic equilibrium between the radiation field and the matter is agood approximation. The typical photon scattering cross section can be takento be roughly the Thomson value σ ≈ 10−24cm2.

(a) How long would it take a photon to random walk from the center to thesurface of the sun in this model?

(b) Estimate the radiant power emitted by the sun, assuming that all energyescapes through the process of photons random walking from the centerof the sun.

(c) Compare the kinetic energy density of the gas to the energy density ofradiation. Find the time taken for the sun to radiate away its internalenergy.

(d) How long could the sun shine with the luminosity estimated in part (b) ifit were powered by hydrogen burning? In this process 4 protons combineto form 4He through a series of strong, weak, and electromagnetic inter-actions. The binding energy of 4He relative to 4 free protons is roughly27.6 MeV. Assume that, over its lifetime, of order ten percent of the sun’smass is available for hydrogen burning.Some useful constants:Radiation density constant a = 7.6× 10−15 erg cm−3 K−4

Boltzmann constant kB = 1.4× 10−16 erg K−1.

Solution

(a) For 1-dimensional random walk the mean square displacement after N scatteringsis given by

〈x2〉 = Nλ2

where N = number of scatterings, and λ is the mean free path.

λ =1

nσ=

1

1.4× 6.02× 1023 × 10−24≈ 1cm

Now in 3D, only 13

of the scatterings contribute to the mean square displacementin a given direction; so radial mean square displacement is

r2 =N

3λ2

Problem 1 from UCLA 1993Spring.

so number of scatterings required for escape is Nes = 3R2

λ2 , and mean timebetween scatterings is λ/c, so escape time is

tes = Nesλ

c=

3R2

λc≈ 5× 1011s ≈ 2× 104yrs.

(b) Radiation energy density ur = aT 4 ≈ 3×1012 erg cm−3, so total radiation energyin the sun is

Ur = (4

3πR3)aT 4 ≈ 4.5× 1045ergs.

Estimate power by assuming Ur is released on time scale tes

P ∼ Urtes∼ 1034erg s−1

(c) The proton density np = ρNA = 8.4 × 1023 cm−3. Thus, the electron plusproton density is n = 1.7 × 1024 cm−3, and the total thermal kinetic energy isapproximately

Uk =4

3πR3(

3

2nkT ) ≈ 2.3× 1048ergs

or Uk ≈ 500Ur. So that it must take about 500 times larger than tes to radiateaway thermal kinetic energy or τ ≈ 3 × 1014s ∼ 107 yrs, the Kelvin-Helmholztime

8 Model of a star

Model a star as an ideal gas in hydrostatic equilibrium, with uniform temper-ature τ = kT. Assume spherical symmetry. (Consider only hydrostatic pressureand neglect other sources, like e.g. radiation pressure.)

(a) Based on equilibrium for a shell, find an equation for the density derivativedρ/dr at radius r, in terms of the gravitational constant G, the density ρ(r)at radius r, mass interior to r, M(r), temperature τ , and gas molecularmass µ. (The relationship between density and number density is ρ = µn.)

(b) Assume the star undergoes a uniform contraction: the distance betweenany two mass elements changes by the same fraction y < 1, so that forevery radius r1 = yr. Find the star’s new temperature τ1 necessary tomaintain hydrostatic equilibrium. Assume the whole surface area of thestar emits radiation as a black body.

(c) How will the color of this star change as it contracts?

(d) How will the luminosity L change as it contracts? (Luminosity is the totalenergy radiated per unit time).

Solution

(a) Consider the pressure and gravitational forces on a shell at radius r:

∆PA = (P (r)− P (r + dr)) 4πr2 = Fg =GM(r)dm

r2=GM(r)ρ(r)4πr2dr

r2

so thatdP

dr= −GM(r)ρ(r)

r2

Substituting the ideal gas law P = nτ = ρτ/µ, we find

dρ

dr= −µ

τ

GM(r)ρ(r)

r2

(b) Rescaling the quantities as r1 = yr, we find ρ1 = ρ/y3 and M1(r1) = M(r). Thusthe new equation of hydrostatic equilibrium is

dρ1dr1

= − µτ1

GM1ρ1r21

.

Tabulating powers of y we find

dρ

dr· y−4 = − µ

τ1

GMρ

r2· y−5.

Thus for consistency we must have

τ1 = τ/y

and the temperature rises as the star contracts.

from Kevin M. Huffenberger

(c) The location of the peak of the Planck spectrum is given by Wien’s displacementlaw,

λmaxT = constant = 2.898× 10−3 m · K,

so that the wavelength of peak emission shrinks as the star contracts and thetemperature rises, as λmax ∝ y. This shifts the color toward the blue end of thespectrum.

(d) The Stefan-Boltzmann law j = σT 4 gives the energy per time radiated per unitarea. So the total luminosity is

L = jA = σT 44πR2

where the radius R provides the surface area of the star. Thus,

L1

L=τ41R

21

τ4R2= y−4y2 = y−2.

1 van der Waals gas 2

Consider a real gas described by the van der Waals equation of state(P +

a

v2

)(v − b) = RT

where P is the pressure, T is the temperature, v is the volume per mole of thegas, R is the universal gas constant, and a and b are constants.

(a) Briefly explain the physical origin and meaning of the constants a and b.

(b) The coefficient of thermal volume expansion is defined as α = v−1(∂v/∂T )P .Show that for the van der Waals gas

α =Rv2(v − b)

RTv3 − 2a(v − b)2.

(c) Show that the isothermal compressibility, κT = −v−1(∂v/∂P )T , for thevan der Waals gas is given by

κT =v2(v − b)2

RTv3 − 2a(v − b)2.

(d) Obtain (∂P/∂T )v and verify that(∂P

∂T

)v

(∂T

∂v

)P

(∂v

∂P

)T

= −1.

Solution

(a) First examine the constant b. In the equation the term v − b takes the place ofthe volume per mole v in the usual ideal gas law Pv = RT . This is because thegas molecules are not pointlike but have a finite size, and so we can interpret bas the volume taken up by a mole of these molecules if they are not moving.

The term involving a actually corrects the ideal gas law, which assumes that themolecules do not interact with each other, and comes from these interactions ina real gas. Since they go down like the volume per mole squared, they actuallygo like 1/r6 where r is the separation of the molecules, and so are of the vander Waals type. Note that the pressure is reduced by the term proportional toa, so each molecule is in fact attracted by the other molecules in the gas whichresults in no net force except near the walls. Near the walls the molecules feela net force away from the wall and so the pressure is reduced.

(b) This is an exercise in finding total differentials. If we are keeping P fixed thenthe total differential of the gas law takes the form

RdT = −2a

v3dv (v − b) +

(P +

a

v2

)dv.

Problem OpaTh-F99-3 from Courses-SC.

Note that the expression we are to find does not involve P , so eliminate P+a/v2

using the gas law

RdT = −2a

v3dv (v − b) +

RT

v − b dv =dv

v3(v − b)(RTv3 − 2a[v − b]2

),

so that

α =1

v

(∂v

∂T

)P

=v2(v − b)

RTv3 − 2a(v − b)2 .

(c) We now do the same again this time keeping T constant and letting P vary, sothat

0 = dP (v − b)− 2a

v3dv (v − b) +

(P +

a

v2

)dv,

and using the same substitution we find

dP (v − b) = − dv

v3(v − b)(RTv3 − 2a[v − b]2

),

so that

κT = −1

v

(∂v

∂P

)T

=v2(v − b)

RTv3 − 2a(v − b)2 .

(d) If v is constant then the above collapses to

dP (v − b) = RdT,

so that (∂P

∂T

)v

=R

v − b ,

and (∂P

∂T

)v

(∂T

∂v

)P

(∂v

∂P

)T

=R

v − b1

vα(−vκT ) = − R

v − bκTα

= −1.

2 Convex lens and concave mirror

An object is 15 cm to the left of a thin convex lens of focal length 10 cm. Aconcave mirror of radius 10 cm is 25 cm to the right of the lens.

(a) Find the position of the final image formed by the mirror and the lens.

(b) Is the image real or virtual, upright or inverted?

(c) Show on a diagram where your eye must be to see this image.

Solution

(a) Careful use the lens equation, which also works for spherical mirrors, and that forthe magnification of these optical systems will yield the results we need. Thelens equation

1

s+

1

s′=

1

f

works for lenses and spherical mirrors, where s is the distance of the object fromthe lens/mirror, s′ is the distance of the image from the lens/mirror, and f isthe focal length which is the image distance when s =∞. These quantities aretaken to be positive when the object, image, or center of curvature lies on thereal side of the optical element. For lenses the real side is the incident side forobjects and the transmission side for images and centers of curvature. Positives′ means a real image, negative s′ means a virtual image. The magnification ism = −s′/s, and if it is negative this means the image is inverted.

Applying this to the lens we have

1

15 cm+

1

s′1=

1

10 cm,

so that s′1 = 30 cm. Note that since s′ is positive this image is real and inverted.Now this image would form 5 cm beyond the surface of the mirror, so it is noton the real side of the mirror. Note that for a spherical mirror the focal lengthis half the radius, so the equation for the mirror looks like

1

−5 cm+

1

s′2=

2

10 cm,

so that s′2 = 2.5 cm. Since this is positive we know that it occurs on the real sideof the mirror inside the focal length, 2.5 cm to the left of the mirror, and so is areal image. Since for the mirror s′2 is positive and s is negative the magnificationis positive so the image remains inverted with respect to the object.

This real image will now behave like an object to the lens, so applying thisequation once more we find

1

22.5 cm+

1

s′3=

1

10 cm,

so that s′3 = 18 cm.

Problem OpaTh-F99-4 from Courses-SC.

(b) This image is on the transmission side of the lens and so is a real image, and isinverted relative to the image from the mirror. The net result is an upright realimage 18 cm to the left of the lens.

(c) Since light rays appear to emanate from this real image but are moving awayfrom the lens, we will need to place our eye to the left of the real image to seeit.

3 Camera

A 35 mm camera has a negative size of 24 mm by 36 mm. It is to be usedto take a picture of a person 175 cm tall in which the image of the person justfills the height (24 mm) of the film. How far should the person stand from thecamera if the focal length of the lens is 50 mm?

Solution

The lens equation gives1

s+

1

s′=

1

f,

and the magnification is m = −s′/s. Using the desired magnification of m = −2.4/175(note it will be negative for a converging lens) we can solve these two equations for s,

1

s+

1

−ms =1

f

1

s

(−m+ 1

−m

)=

1

f

s = f

(−m+ 1

−m

)= 5 cm

(2.4/175 + 1

2.4/175

)= 369.6 cm.

Problem OpaTh-F99-5 from Courses-SC.

4 Cross section

Consider scattering in the central potential (r = |~r|)

U =(u0r

)exp

[−( ra

)](1) Calculate the differential cross section in the first Born approximation

dσ

dΩ=

m2

4π2h4

∣∣∣∣∫ d3rU exp (−i~q · ~r)∣∣∣∣2

where ~q = ~kout − ~kint is the difference of the wave vector of the incomingand outgoing particle.

(2) Find the total cross section σtot (k). Hint: q = |~q| is related to the scatteringangle θ by q = 2k sin θ

2

Solution

(1) We use spherical coordinates with the z-axis in ~q direction. The dφ integrationgives 2π. So, with x = cos θ (this is not the scattering angle)

∫d3r

rexp

[−( ra

)− i~q · ~r

]= 2π

∫ ∞0

rdr

∫ −1

+1

dx exp[−( ra

)− (iqr)x

]=

2πi

q

∫ ∞0

dr

exp[−( ra

)+ iqr

]− exp

[−( ra

)− iqr

]=

2πi

q

∫ ∞0

dr exp [−c−r]− exp [−c+r]

=2πi

q

(1

c+− 1

c−

)where c± = a−1 ± iq = a/ (1 + iaq). Now,

2πi

q

(1

c+− 1

c−

)=

4πa2

1 + a2q2.

Therefore,(dσ

dΩ

)=m2u2

0

4π2h4

∣∣∣∣∫ d3r

rexp

[−( ra

)− i~q · ~r

]∣∣∣∣2 =m2u2

0a4

h4 (1 + a2q2)2

(2) As dΩ = dφd cos θ, the dθ integration is trivial (= 2π) and the relevant integral is(with y = sin2 (θ/2) and using cos (θ) = cos2 (θ/2)−sin2 (θ/2) = 1−2 cos2 (θ/2))

∫ +1

−1

d cos θ[1 + 4k2a2 sin2 (θ/2)

]2 = 2

∫ 1

0

dy

[1 + 4k2a2y]2= − 1

2k2a2 (1 + 4k2a2y)

∣∣∣∣10

= − 1

2k2a2 (1 + 4k2a2)+

1

2k2a2=−1 +

(1 + 4k2a2

)2k2a2 (1 + 4k2a2)

=2

1 + 4k2a2

The total cross section is then

σ =

∫dσ =

4m2u20a

4

h2 ·∫

dΩ[1 + 4k2a2 sin2 (θ/2)

]2 =16πm2u2

0a4

h4 (1 + 4k2a2)

5 Ideal gas in a cycle

An ideal gas (γ = 1.5) follows the cycle shown in the figure.

(a) If P1 = 1 atm, V1 = 100 L, and T1 = 20 C, find the amount of gasconsidered at point 1.

(b) The gas is then heated at constant volume to point 2, where the pressureis P2 = 2 atm. What is the temperature of the gas there?

(c) Next, the gas is isothermally expanded to point 3, such that P3 = P1 = 1atm. What is the work W23 done during the isothermal expansion?

(d) How is the result modified if an adiabatic rather than isothermal expansionis considered between points 2 and 3?

Solution

(a) Use the ideal gas law and the value in the appropriate unitsR = 0.0821 L · atm/(mol ·K),and using T1 = 293 K, we have

n =P1V1

RT1= 4.16 mol.

(b) Using the ideal gas law again, we have that since V is constant

P1

T1=

P2

T2

T2 =P2

P1T1 = 586 K.

(c) Let’s first find V3. Since the temperature is unchanged from T2, the volume isgiven by the ideal gas law

P3V3 = P2V2 = P2V1

Problem OpaTh-F99-6 from Courses-SC.

to be

V3 =P2

P3V1 =

P2

P1V1 = 2V1 = 200 L.

For an isothermal expansion we have

W2→3 =

∫ V3

V1

P dV

and using the ideal gas law we have that P = nRT2/V , so that

W2→3 = nRT2

∫ V3

V1

1

VdV = nRT2 ln(V3/V1)

= (4.16 mol) · (0.0821 L · atm/(mol ·K)) · (586 K) ln(2) = 139 L · atm.

(d) For an adiabatic expansion we have no heat added to the system, and the expan-sion follows

PV γ = constant,

where γ = Cp/Cv = 1.5. The volume is now changed to

V ′3 =

(P2

P3

) 1γ

= 22/3V1.

The work integral is modified to

W2→3 = P2Vγ2

∫ V ′3

V1

1

V 3/2dV = P2V

3/21

[−2V −1/2

]V ′3

V1

= 2P2V3/21

(V−1/21 − 2−1/3 · V −1/2

1

)= 2P2V1

(1− 2−1/3

)= 2(2 atm) · (100 L)

(1− 2−1/3

)= 82.5 L · atm.

6 Two lenses

Two lenses are separated by 35 cm. An object is placed 20 cm to the left ofthe first lens, which is a converging lens of focal length 10 cm. The second lensis a diverging lens of focal length −15 cm. What is the position of the finalimage? Is the image real or virtual? Upright or inverted? What is the overallmagnification of the image?

s2

s

s s

f1

1

1

2f

2

Solution

We can find everything we need from the lens formula(e)

1

s+

1

s′=

1

f, m = −s

′

s,

being careful to use a negative f for the diverging lens (since the focal point of lighton the incident side is on the incident side and not the transmitted side, it is thereforenegative), and use the figure to check our answers. For the first lens we have

1

s′1=

1

f1− 1

s1=

1

10 cm− 1

20 cm=

1

20 cm,

so that the first real image is 20 cm from the lens and has

m1 = −s′1

s1= −1,

so that the image is the same size as the object and inverted.

For the second lens we have

1

s′2=

1

f2− 1

s2=

1

−15 cm− 1

15 cm= − 2

15 cm,

so that the image is virtual (it is on the incident side, since s′2 < 0) and 7.5 cm to theleft of the second lens. The magnification of the second lens is

m2 = −s′2

s2= −−7.5 cm

15 cm=

1

2,

Problem OpaTh-F99-7 from Courses-SC.

so that the combined magnification is m = m1m2 = −0.5 and the virtual image is

inverted with an overall magnification of −0.5.

7 Cylinder with piston

The cylinder shown in the figure has a piston of mass M that can slide withoutfriction. The area of the piston is S and the cylinder is filled with an ideal gas(γ = 1.5), with an initial volume V , and an initial pressure P . Assume that theoutside pressure on the piston is zero (vacuum).

(a) Determine the initial acceleration of the piston.

(b) Calculate the velocity of the piston after it has moved a distance L, as-suming that the gas is thermally isolated (adiabatic).

(c) What distance would the piston have to move for the temperature of thegas to drop to one half of the original value?

Solution

(a) The initial force on the piston is given by the initial pressure P times the areaof the piston to which this pressure is applied, F = P · A. Then the initialacceleration is

a =F

M=PA

M.

(b) Adiabatic expansions follow the rule PV γ = constant, where γ is the ratioCP /CV . We can find the constant involved by using the initial condition, sothat at some later time when the pressure is P ′ and the volume V ′ we have

P ′V ′γ = PV γ ,

so that

P ′ = P

(V

V ′

)γ.

In order to find the velocity of the piston we will use conservation of energy,which states that the work done by the gas during the expansion is equal to thekinetic energy gained by the piston. The work done by the gas is

W =

∫ Vf

V

P ′ dV ′ = PV32

∫ Vf

V

1

V ′32

dV ′ = PV32

[−2

V ′12

]VfV

= 2PV

[1−

(V

Vf

) 12

].

Problem OpaTh-F99-8 from Courses-SC.

The final volume is Vf = V + LS so that

W = 2PV

[1−

(V

V + LS

) 12

],

and using W = Mv2/2 we have v = (2W/M)1/2 and

v =

(4PV

M

[1−

(V

V + LS

) 12

]) 12

.

(c) We will combine the ideal gas law with the adiabatic expansion law. Since theideal gas law gives P = nRT/V then we have that

PV γ = nRTV γ−1 = nRTfVγ−1f ,

so that

TfT

=

(V

Vf

)γ−1

=

√V

Vf=

√V

V + LS=

1

2,

so that4V = V + LS

and

L =3V

S.

8 Wedge of air

A wedge of air is formed between two glass plates held apart at one edge by asheet of paper whose thickness is 4.1×10−5 m. Green light (λvacuum = 552 nm)strikes the glass plates nearly perpendicularly. Assume nglass = 1.52, nair =1.00.

(a) How many bright fringes occur between the place where the plates touchand the edge of the sheet of the paper?

(b) Is there a dark fringe or a bright fringe where the plates touch? Why?

Solution

(a) The fringes occur because of interference between reflected rays 1 and 2 as illus-

trated in the diagram. Ray 1 is not inverted at the glass/air interface becauseit is going from a more dense to a less dense medium. Transmitted waves arealways transmitted in phase. The reflected ray 2 is inverted at the air/glassboundary, so the two reflected rays are out of phase by π before we take intoaccount the path difference. We will get bright fringes when the path differenceδ is a half-integral number of wavelengths, so that

δ = (m+1

2)λair, m = 0, 1, 2, ....

At the right edge the path difference is just twice the thickness t of the paper,so there

2t = (m+1

2)λair,

and we see that the last fringe has

m =2t

λair− 1

2=

2 · 4.1× 10−5

552× 10−9− 1

2= 148,

and since we start counting from zero we see that we have 149 bright fringes.

(b) As we saw above, if there is no path difference the two reflected rays have a phasedifference of π and destructively interfere. This means that close to the contactpoint we are in a dark fringe.

Problem OpaTh-F99-9 from Courses-SC.

9 Ideal gas in a cycle 2

In the cycle shown in the figure, 1 mol of an ideal gas (γ = 1.4) is initially ata pressure of 1 atm and a temperature of 0C. The gas is heated at constantvolume to T2 = 150C amd is then expanded adiabatically until its pressureis again 1 atm. It is then compressed at constant pressure back to its originalstate. Recall R = 0.082 L · atm/(mol ·K). Find:

(a) the temperature T3 after the adiabatic expansion

(b) the heat entering or leaving the system during each process

(c) the efficiency of this cycle

(d) the efficiency of a Carnot cycle operating between the temperature extremesof this cycle.

Solution

(a) To find the temperature T3 we will need the volume V2 = V1 and and the pressureP2 before the expansion, which we can find using the ideal gas law. We have

V1 =nRT1

P1=

1 mol · 0.082 L·atmmol·K · 273 K

1 atm= 22.4 L,

and so

P2 =nrT2

V1=

1 mol · 0.082 L·atmmol·K · 423 K

22.4 L= 1.55 atm.

Now since the expansion is adiabatic to pressure P1 we can find V3 using therule that PV γ = constant, so that

V γ3 =P2

P3V γ1 ,

Problem OpaTh-F99-10 from Courses-SC.

and so

V3 =

(P2

P3

)1/1.4

V1 = (1.55)1/1.4 · 22.4 L = 30.6 L.

Now we can find T3 using the ideal gas law,

T3 =P1V3

nR=

1 atm · 30.6 L

1 mol · 0.082 L·atmmol·K

= 373 K.

(b) Since there is no work done during 1 → 2 the heat entering the system is thechange in the internal energy of the system, which is

Q1→2 = CV (T2 − T1).

We are not given the value of CV , but since γ = 1.4 = CP /CV = 1 + nR/CVwe know that CV = 5nR/2, and so

Q1→2 =5

2nR(150 K) = 30.75 L · atm.

Since the expansion 2→ 3 is adiabatic, no heat enters or leaves the system. Forthe compression 3→ 1 at constant pressure P1 we can use the definition of theheat capacity at constant pressure

Q3→1 = CP (T3 − T1) =7

2nR(−100 K) = −28.7 L · atm.

(c) The efficiency of the cycle is defined to be the net work done during the cycledivided by the heat added during the phase when we add heat, which is 1→ 2,so that

ε =W

Q1→2.

Note, however, that since the internal energy of the gas does not change after acomplete cycle, we must have that W = Q1→2 +Q3→1, so that

ε = 1 +Q3→1

Q1→2= 1− 28.7

30.75= 6.7%.

(d) An ideal Carnot cycle which operates between the temperature T2 = Thot and thetemperature T3 = Tcold (the maximum temperature during the heating phaseand the minimum temperature during the cooling phase, respectively) wouldhave an isothermal expansion at the hot temperature Thot and an isothermalcompression at the low temperature Tcold, with the other two processes be-ing an adiabatic compression (to Thot) and an adiabatic expansion (to Tcold).Isothermal expansions and compressions involve heats Qhot and Qcold which areproportional to the temperature, through

Q = nRT ln

(VfVi

).

Once can show using the rule for adibatic expansions that the ratios of thevolumes at the beginning and end of the adiabatic processes are equal, so that theratio of the two heats is simply the ratio of the temperatures and the efficiencyof the Carnot cycle is simply

ε = 1 +Qcold

Qhot= 1− Tcold

Thot= 1− 273

423= 35.5%.

10 Ideal gas in a cycle 3

The figure shows three reversible processes joining initial and final states of nmoles of a monatomic gas. For the three processes calculate:

(a) the work done on the system

(b) the heat interchanged with the environment

(c) the change in internal energy

(d) the change in entropy. Express all results in terms of n and the initialtemperature T0.

Solution

(a) In what follows we will calculate the work done by the system (the usual con-vention) so that the answer to the question is always the negative of this.

Label the path from i to f which uses an expansion at constant pressure as path1; it does work

W1 =

∫P dV = P0

∫dV = P0V0 = nRT0.

Label the diagonal path as 2. To find the work done along this path we needto know Pf , which we can find from the ideal gas law, since we know from thethird (curved) path that the final temperature has to be T0. Hence

Pf =nRT0

2V0=P0V0

2V0=P0

2,

Problem OpaTh-F99-11 from Courses-SC.

so that (using geometry to find the area)

W2 = V0P0 + P0/2

2=

3

4P0V0 =

3

4nRT0.

Finally, the third path is an isothermal expansion where P = nRT0/V so that

W3 =

∫ 2V0

V0

nRT0

VdV = nRT0 ln(2) = nRT0 ln(2).

(b) The heat added to the system is the work W done by the system plus the changein the internal energy of the system. For all paths the internal energy of thesystem does not change since the initial and final temperatures are the same, soin all cases the heat added to the system is simply the work done by the system.

(c) Zero in all cases since the initial and final temperatures are equal.

(d) The entropy changes by dS = dQrev/T when an amount of heat dQrev is addedat a temperature T in a reversible process. For an ideal gas we have

dS =dU

T+dW

T= CV

dT

T+ P

dV

T= CV

dT

T+ nR

dV

V,

so that

∆S = CV

∫ Tf

Ti

dT

T+ nR

∫ Vf

Vi

dV

V.

Since for all paths Ti = Tf we have that

∆S = nR

∫ Vf

Vi

dV

V= nR

∫ 2V0

V0

dV

V= nR ln(2).

1 Concepts 3

Provide a brief qualitative description for each item listed below.

(a) Compton effect.

(b) the difference between bosons and fermions.

Solution

(a) Compton scattering is the process of the scattering of a photon by an electron(although it can also happen for other charged particles). The collision conservesenergy and momentum, so the electron recoils and the photon is emitted, ingeneral, at an angle to its original direction and with a reduced energy. The neteffect is that the photon appears to scatter from the electron with a reducedfinal energy and so longer wavelength. The change in the wavelength dependsonly on the scattering angle and not on the original wavelength.

(b) Bosons are particles like the photon which have integral spin (or intrinsic angularmomentum), while fermions are particles like the electron that have half-integralspin. Fermions satisfy Fermi-Dirac statistics, because they obey the Pauli exclu-sion principle, which states that no two indistinguishable fermions can occupythe same quantum mechanical state. Bosons satisfy Bose-Einstein statistics,which means that they do not obey the Pauli exclusion principle.

Problem ModQM-F99-8 from 2003-Proficiency-Spring.

2 Concepts 4

Provide a brief qualitative description for each item listed below.

(a) Hyperfine structure in a hydrogen atom.

(b) Rayleigh scattering.

Solution

(a) The proton that makes up the hydrogen nucleus has a spin, and since it is acharged particle it has a magnetic moment. The same is true of the electron.These two magnetic moments interact with each other (one moves in the mag-netic field generated by the other) and the resulting interaction Hamiltoniancause a splitting of the ground state (and other energy levels) which depends onthe projection of the spin of the electron. The energy difference is very small,of the order of 10−7 eV, and wavelength of the light from this transition is 21cm, in the radio part of the spectrum, which is important for radio astronomy.

(b) Rayleigh scattering is a single-step process where a photon is absorbed by anatom and promotes an electron into an excited state, which then decays back tothe ground state. The incident and scattered photons are correlated, and havethe same energy.

Problem ModQM-F99-9 from 2003-Proficiency-Spring.

3 Concepts exp obs

For the following pivotal experiments and observations in physics and astron-omy, describe the key observation and why it changed our understanding of thephysical world:

(A) Blackbody radiation spectrum

(B) Photoelectric effect

(C) Compton scattering

(D) Rutherford scattering

(E) 2011 Nobel prize in physics to Saul Perlmutter, Brian Schmidt, and AdamRiess

Solution

(A) The key observation was the peaking of the EM intensity at a temperature-dependent wavelength and its decrease at both shorter and longer wavelengths.Max Planck’s equation which reproduced the effect formed the foundation ofquantum mechanics by implying that light is only emitted in quanta of energyhf .

(B) Key observation: The kinetic energy of each electron released from a metal surfacedepends only on the wavelength of the light absorbed, not on its intensity. AlbertEinstein interpreted this to show that light is only absorbed in quanta of energyhf . This provided a major confirmation of quantum mechanics.

(C) Key observation: The kinematics of the scattering of quanta of light off electronsdemonstrated the particle nature of light and confirmed the wave-particle dualityat the core of quantum mechanics.

(D) Ernest Rutherford’s observation that alpha particles could scatter backward offgold atoms negated the Thompson plum pudding model of the atom and led tothe understanding of the atom as a tiny, heavy nucleus surrounded by a cloudof light electrons.

(E) Their careful measurement of the Hubble graph (distance vs. recession velocity)of distant Type 1A supernovae revolutionized cosmology by showing that therate of expansion of the universe is increasing. This in turn implies that thedominant energy of the universe is repulsive.

4 Photoelectric effect 2

Monochromatic blue light with a wavelength of 434.2 nm is incident on asample of cesium. Electrons emitted from the cesium surface are observed tohave velocities ranging up to 5.491 × 105 m/s. Note me = 0.511 MeV/c2,qe = 1.602 × 10−19 C, hc = 1240 eV·nm.

(a) What is the work function for this sample of cesium?

(b) Explain why there is a range of emitted electron velocities.

(c) What is the wavelength of the fastest emitted electrons?

(d) Now assume that the hydrogen discharge lamp produces 2.0 µW of powerradiated in this particular blue Balmer series spectral line. If the lampcan be considered to be a point source and emits the light isotropically,estimate how many of these blue photons per second strike a circularcesium sample 7.5 cm in diameter and placed 10 cm from the lamp.

Solution

(a) The maximum kinetic energy of the emitted electron plus the work function ofthe cesium is equal to the energy of the photon. This means that we can findthe work function by

φ = Eγ − Tmax = hν − Tmax =hc

λ− 1

2mev

2max

=1240 eV · nm

434.2 nm− 1

2· 0.511× 106 eV · (5.491× 105

3.0× 108)2

= 2.0 eV.

(b) Electrons near the surface absorb all of the energy of the incident photon andthen leave the surface with the maximum energy available, which is the photonenergy minus the work function. Electrons deeper inside the metal are morelikely to interact with the electrons around other atoms on their way out of thesample before leaving the surface, and these interactions will reduce their kineticenergy.

(c) The de Broglie wavelength is

λ =h

p=

hc

mvc=

1240 eV · nm

0.511× 106 eV(5.491× 105/3.0× 108)= 1.33 nm.

(d) It is actually a difficult geometrical problem to find the area cut out of a sphereby a circular disk which has its circumference touching the inside of the disk.We need this to find the solid angle subtended by the cesium sample and so thefraction of the emitted energy which is absorbed by the sample. If the radius ofthe disk is small compared to that of the sphere we can approximate the solidangle by the area of the disk divided by the surface area of the sphere whichgoes through the center of the disk, but that is not the case here. This problem

Problem ModQM-F99-10 from 2003-Proficiency-Spring.

was not intended to be that hard, so we will use this approximation even thoughit is not justified. The solid angle is, with this approximation,

Ω =πr2

4πR2=

1

4

(7.5/2

10.0

)2

= 0.0351

and so the energy E per second which goes into this solid angle is this fractionof the total emitted energy per second,

E = 0.0351·Ptot·1 s = 0.0351·2×10−6 J =0.0702× 10−6

1.602× 10−19 J/eV= 4.38×1011 eV,

and the number of photons per second is this energy divided by the energyhν = hc/λ per photon,

N =2.19× 1011 eV

1240 eV · nm/434.2 nm= 1.53× 1011 photons.

5 Compton scattering

A photon with wavelength 24.8 fm strikes a proton at rest. The photonundergoes Compton scattering, and the scattered photon is seen by an observerin the lab to be emitted at 180 with respect to the direction of the incidentphoton. The mass of the proton is Mp = 938 MeV/c2, and hc = 1240 MeV·fm.

(a) What is the energy of the incident photon? What name would typically begiven to classify this “type” of photon? Give a very brief explanation foryour choice.

(b) Using relativistic kinematics, find (i) the wavelength of the scattered photonand (ii) the de Broglie wavelength of the recoiling proton.

(c) If we could observe this reaction occuring in the center-of-mass frame in-stead of the lab frame, what would we then see as the difference betweenthe wavelengths of the incoming and scattered photons? Explain youranswer.

Solution

(a) The energy of the photon is

E = hν =hc

λ=

1240 eV · nm

24.8× 10−6 nm= 50 MeV,

which is a gamma ray. Gamma rays are typically emitted during transitionsbetween excited states and ground states of nuclei, and typical nuclear excitationenergies are in the MeV range.

(b) We have to conserve relativistic energy and momentum in the collison. The initialenergy and momentum of the photon are related by Eγ = pc, or we can writethat the photon four-momentum is (assume all motion is along the z-axis)

pγ = (Eγ ,pγ) = (pc, 0, 0, p),

while the initial proton four momentum is

P = (mpc2, 0, 0, 0).

The final photon four momentum is

p′γ = (p′c, 0, 0,−p′),

since the photon back scatters, and by conservation of four-momentum (i.e.energy and momentum) we know the final four momentum of the proton lookslike

P ′ = (mpc2 + pc− p′c, 0, 0, p+ p′).

This takes care of conservation of energy and momentum, and we can now usethis to find p′ in terms of p if we use that the square of the proton four-momentumis

P ′2 = m2pc

2

Problem ModQM-F99-11 from 2003-Proficiency-Spring.

which is equivalent to the relation

E′2 = P′2c2 +m2pc

4.

This gives the relation

m2pc

4 = (mpc2 + pc− p′c)2 − (p+ p′)2c2

= m2pc

4 + 2mpc3(p− p′) + (p− p′)2c2 − (p+ p′)2c2

4pp′c2 = 2mpc3(p− p′)

2pp′ = mpc(p− p′)

p′ =mpcp

mpc+ 2p

p′c =mpc

2pc

mpc2 + 2pc= 50 MeV · 938

938 + 100= 45.2 MeV

so that the photon wavelength is

λ′ =h

p′=hc

p′c=hc

pc

(mpc

2 + 2pc

mpc2

)= λ · 938 + 100

938= 27.4 fm.

Now we know that the momentum of the proton is p+p′, so that the de Brogliewavelength of the proton is

λp =hc

pc+ p′c=

1240 eV · nm

(50.0 + 45.2) MeV= 13.0 fm.

General notes The energy momentum conservation for relativistic four vectorsis

P + pγ = P ′ + p′γ . (1)

Since the angel between initial and final photons is known and nothing is knownabout the direction of the final proton it is convenient to rewrite this equationas

P ′ = P + pγ − p′γ , (2)

and then take a relativistic square of both sides

m2p = m2

p + 2Ppγ − 2Pp′γ + 2pγp′γ . (3)

This leads to Compton’s equation

ω′ =m

mp + ω(1− cos θ)(4)

for the frequency of the scattered photon.

(c) In the center of mass frame the initial momentum of the photon and proton mustbe equal and opposite, and the same is true of the final momenta. Then theconservation of energy condition looks like

pc+√m2pc4 + p2c2 = p′c+

√m2pc4 + p′2c2.

Since the left and right sides of this equation are the same function of p (or p′)then this can only be true if p = p′, so the incident and scattered photons musthave the same momentum.

6 Decay of tritium 2

A vessel holds 2 µg of tritium.

(a) What is the initial decay rate of the tritium?

(b) How much time will elapse before the amount of tritium falls to 1% of itsinitial value?

Solution

(a) The number N of tritium nuclei as a function of time is

N = N0e− ln(2) t/t1/2 ,

so that the rate of decay is

dN

dt= N0

(− ln(2)

t1/2

)e− ln(2) t/t1/2 ,

which at time t = 0 isdN

dt(0) = −N0

(ln(2)

t1/2

).

The total number of nuclei is the mass divided by the mass per nucleus, and thehalf life is 12.3 y, so that

dN

dt(0) = − 2× 10−9 kg

5.01× 10−27 kg

(ln(2)

12.3 · 365 · 24 · 3600 s

)= −7.13× 108 s−1.

(b) If N/N0 = 1/100, then we have that

e− ln(2) t/t1/2 =1

100,

or by taking the natural log of both sides

ln(2)t

t1/2= ln(100),

and so

t =ln(100)

ln(2)t1/2 = 81.7 y.

Problem ModQM-S00-2 from 2003-Proficiency-Spring.

7 Interference of coherent light

Four equally spaced coherent light sources with wavelength of 500 nm areseparated by a distance of d = 0.1 mm. The interference pattern is viewed ona screen at a distance of 1.4 m. Finf the positions of the principal interferencemaxima and compare their width with that for just two sources with the samespacing.

Solution

Start by looking again at the case of two light sources. In this case when the pathdifference d sin(θ), where θ is the angle from the normal to the plane of the light sources,is an integral number of wavelengths we have constructive interference and so aninterference maximum. If this path difference is a half-integral number of wavelengthswe have destructive interference and so an interference minimum. This means thatthe interference maxima are at angular positions

sin(θ) = mλ

d(maximum, two sources)

and the minima are at

sin(θ) =

(m+

1

2

)λ

d(minimum, two sources),

so that the maxima have angular width equal to their angular spacing, λ/d.

In the case of four sources, the path difference between adjacent sources is still d sin(θ),and the condition for a principal interference maximum (intensity 16 times that of asingle slit) is the same, but the condition for an interference minimum changes. Thereis obviously an interference maximum for d sin(θ) = 0; the first minimum will occurwhen the four phasors from each source form a square (see diagram), and so the phaseangle between adjacent phasors is 90. This means that we have

d sin(θ) =λ

4(first minimum, four sources).

By the same idea there are obviously interference minima when d sin(θ) = λ/2 and3λ/4. In between these angles we have secondary maxima (where the intensity is 1/16times that of the principal maxima). The next principal maximum occurs when thephase angle between each phasor is again zero, so d sin(θ) = λ.

This means that the principal interference maxima are at angles

sin(θ) = mλ

d, m = 1, 2, 3...

Problem S00Prof-11 from Courses-SC.

and have half the width that they have in the case of two sources. Converting this todistances we can use the small angles approximation sin(θ) = θ to write that positionsof the principal maxima

yn = L sin(θn) ' Lθn = Lnλ

d= (1.4 m)

n · 5× 10−7 m

1× 10−4 m= n(7 mm),

and so the principal maxima are separated by 7 mm, and their width is one half of

this distance, 3.5 mm. For two sources this width would be 7 mm.

8 Photoelectric effect 3

During a photoelectric effect experiment, sodium metal is illuminated withlight of wavelength 4.20×102 nm. The stopping potential is found to be 0.65 V.When the wavelength is changed to 3.10×102 nm, the stopping potential isfound to be 1.69 V. Using only these data and the values of the speed of light,c = 3.00× 108 m/s, and the elementary charge, e = 1.60× 10−19 C, find a valuefor Planck’s constant.

Solution

Recall that in such experiments, photons are absorbed by the conduction electrons inthe surface of the metal, which then have enough kinetic energy that they are no longerbound to the metal. The minimum energy required to remove an electron from thesurface (which is then moving with zero velocity and so no kinetic energy) is the workfunction φ. This is measured by detecting a current flowing out of the surface througha vacuum into an electrode, where a voltage is applied which drives the electrons backinto the metal’s surface. When this voltage reaches the largest kinetic energy theelectrons have been given from the photons the current stops.From this and conservation of energy we know that

hν = φ+ eV,

where V is the stopping potential. If we have two frequencies ν and two values of thestopping potential we can eliminate the work function by writing

hν1 − eV1 = hν2 − eV2,

so thath(ν1 − ν2) = hc(1/λ1 − 1/λ2) = e(V1 − V2).

Solving for h we find

h =e(V1 − V2)

c(1/λ1 − 1/λ2)

=1.6× 10−19 C · (0.65− 1.69) V

3.0× 108 m/s · (1/4.2× 10−7 − 1/3.10× 10−7) m−1

= 6.6× 10−34 J · s.

Problem ModQM-F00 from Courses-SC.

9 Beta decay

In nuclear beta decay, electrons are observed to be ejected from the atomicnucleus. Assume that electrons are somehow trapped within the nucleus andthat occasionally one escapes.

(a) Estimate the kinetic energy that such an electron must have. Assume anuclear diameter of 1.0 × 10−14 m.

(b) Electrons emitted from the nucleus in nuclear beta decay typically havekinetic energies of about 1 MeV. Comment on the difference between theactual kinetic energy and your result for part a.

Solution

(a) The statement of this problem is confusing. In beta decay electrons are ejectedforcibly from the nucleus, since they actually come from beta decay of a neutroninto a proton and an electron (and an anti-neutrino), which releases energy. Apart of this energy is in the form of rest and kinetic energy of the electron,another (smaller) part goes into the recoil energy of the proton and the neutrinoenergy. Since for a free neutron about 1.3 MeV is available in total for this,and we need 0.5 MeV for the electron rest energy, one might guess that in theabsence of all other effects the kinetic energy of the electron would be veryroughly one MeV. However, one also has to take into account the difference inthe nuclear binding energy of the initial and final nuclei, and nuclear structureeffects (the initial neutron may need to be in a higher energy level than the finalproton because of the Pauli principle and the fact that large nuclei have moreneutrons than protons). In addition, the ejected electron is attracted to thepositive charge of the nucleus and so a (possibly substantial) part of its initialkinetic energy is required to overcome this Coulomb barrier if it is to becomefree.

What we are asked to do is to suspend all of our knowledge of the physics ofthis process and imagine an electron which is confined within the nucleus allof the time. Since we are not given anything except the size of the nucleus, inparticular we do not know its charge, we cannot evaluate the energy requiredto overcome the Coulomb barrier. If all we know is the size of the (roughlyspherical) nucleus which contains the electron, then all we can do is to estimateits momentum and so kinetic energy using the uncertainty principle.

If the electron is confined to a region of size 10−14 m, then by the uncertaintyprinciple it must have a ∆x smaller than this, but how much smaller? Thisis the essential problem with using the uncertainty principle for doing physics.I got curious about this and decided to find ∆x for a particle in its groundstate in a one-dimensional box x ∈ [−a, a], which has wavefunction ψ(x) =(1/√a) cos(πx/2a). The result is that

⟨x2⟩

=1

a

∫ +a

−ax2 cos2(πx/2a) = a2

(π2 − 6)

3π2,

Problem ModQM-F00-3 from Courses-SC.

which since 〈x〉 = 0 gives

∆x =√〈x2〉 =

a

π

√π2 − 6

3π2= 0.361 a.

So if, as is usually done, we say that the uncertainty ∆x in the electron’s positionis the width 2a of the box, we make a mistake of a factor of 2a/(0.361a) = 5.5in ∆x, and so greatly underestimate ∆p. Perhaps this approach to estimatingparticle energies should be called the ‘uncertain principle’, good for an order ofmagnitude ar best.

If we forge ahead regardless of this problem and use the diameter d of the nucleusfor ∆x then we will very roughly estimate the momentum of the electron in thenucleus by using

∆p ≥ h/2

∆x=

h

2d=

hc

2dc=

197.32 (MeV/c) · fm20 fm

= 9.85MeV

c.

where we have used the convenient unit hc = 197.32 MeV·fm and that thenuclear diameter given is 10 fm (1 fm=10−15 m).

Since pc is larger (by a factor of 20) than the rest energy 0.511 MeV of theelectron we know that this is a relativistic electron, and we have to use therelation (assume, again probably erroneously, that we can use ∆p for the r.m.s.momentum)

E = (p2c2 +m2c4) = 9.86 MeV ' pc.

(b) See all the physics reasons above for why this is bound to be a poor estimate ofthe energy of a beta-decay electron.

10 Double-slit interference

Two slits of width a = 0.015 mm are separated by a distance d = 0.06 mmand illuminated by light of wavelength λ = 650 nm. How many bright fringesare seen in the central diffraction maximum?

Solution

The condition for a minimum of the diffraction pattern is

a

2sin(θ) =

nλ

2, n = 1, 2, 3...,

where a is the slit width. The condition for a maximum of the interference pattern is

d sin(θ) = mλ, m = 0, 1, 2, ...,

where d is the slit spacing. Since in this case d = 4a, we have that the m = 4

interference maximum occurs at the position of the n = 1 diffraction minimum. That

means that there are only three interference maxima on either side of the central

diffraction, not counting the central (m = 0) maximum, since the fourth maximum

will be wiped out by the diffraction pattern. This gives seven bright interference fringes

within the central diffraction maximum, regardless of the wavelength of the light.

Problem OpaTh-F99-2 from Courses-SC.

1 Definitions

Define the following properties of a solid, and explain how each can be measured

(a) electrical resistivity

(b) magnetic susceptibility

(c) specific heat

(d) thermal conductivity

(e) dielectric constant

Solution

(a) Electrical resistivity: ρ, the constant of proportionality that relates R = ρ(L/A),where L is the length of a wire, A is the cross-sectional area, and R is theresistance. To measure the resistivity of a conductor, you would measure thelength of the wire, the cross-sectional area, and then the resistance throughV = IR or R = V/I (V and I are the voltage and current, respectively).

(b) Magnetic susceptibility: χ = M/B, where M is the magnetization, and B is theapplied magnetic field. The magnetic susceptibility can be measured in sev-eral different ways - the Gouy balance is one method, where you can measurethe torque in a magnetic field of a material. The magnet remains stationarywhile the sample moves, giving an apparent weight loss or gain. Another wayis through SQUID magnetometry (Superconducting Quantum Interference De-vice), which measures changes in the magnetic flux quantum.

(c) Specific heat: the amount of heat required to raise the temperature of either 1 gor 1 mol of a substance by 1 K: c = (1/m) ∗∆Q/∆T (m is the mass or molarmass, Q is the amount of heat, T is the temperature). You can measure thespecific heat by adding heat to a known amount of substance and then measurethe temperature change. Of course, everything must be well insulated to preventexcess heat loss or gain to the sample.

(d) Thermal conductivity: the constant of proportionality k that relates the heat flowin a certain amount of time to the value of the heat gradient across an object ofcross-sectional area A.

∆Q

∆t= −kA∆T

∆x(1)

Here, ∆Q is the amount of heat that flows in time ∆t, the rod has cross-sectionalareaA, and the gradient is ∆T/∆x. To measure this, you would need to establisha temperature gradient across a cylindrical object that you wish to measure (oflength x and cross-sectional area A), and then you would need to measure theheat flow as a function of time. These experiments are done in thermally isolatedcontainers to prevent heat loss from other sources.

Problem 9 from Qualifying exam 2007-spring.

(e) Dielectric constant: this is the factor by which the internal electric field of amaterial changes when an electrical field is applied: κ = εs/ε0 where εs isthe static permittivity and ε0 is a constant. This can be measured through acapacitor, where the capacitance is C = κε0A/d (A is the cross-sectional areaand d is the distance between the plates). In metals, the dielectric constant canbe infinite, and so other methods are needed.

2 Ideal gas in a cycle 5

Consider an ideal monatomic gas consisting of N particles each with mass m.The gas is first expanded from volume V1 to volume V2 at constant pressure sothat P1 = P2. The gas is then further expanded isothermally to a final volumeV3 with pressure P3. The corresponding temperatures are T1, T2 and T3, withT2 = T3.

(a) Using classical thermodynamics compute the total entropy change andexpress it in terms of V1, V3, T1, T3. (Recall that the heat capacity ofan ideal monatomic gas at constant pressure is CP = 5

2NkB , where kB isBoltzmann’s constant).

(b) Using the partition function of the ideal classical monatomic gas, calculatethe Helmholtz free energy of the system as a function of particles N , thevolume V , and the temperature T .

(c) Using the Helmholtz free energy obtained in part (b), obtain an expressionfor the entropy of the system. Use this expression to compute the totalentropy change of the gas and compare your result to that obtained inpart (a).

Solution

(a) For an ideal monatomic gas CV = 32NkB and CP = 5

2NkB . For the constant

pressure expansion the change of the entropy of the gas is

∆S1→2 =

∫ T2

T1

CPTdT = CP ln

T2

T1=

5

2NkB ln

T2

T1. (2)

For the isothermal expansion the internal energy of the gas does not change.This means the heat added is equal to the work done by the gas.

∆Q2→3 =

∫ V3

V2

PdV =

∫ V3

V2

NkBT

VdV = NkBT ln

V3

V2. (3)

Thus the change in entropy is

∆S2→3 =∆Q2→3

T= NkB ln

V3

V2. (4)

Using the fact that T2 = T3, and V2 = V1T2/T1 we have for the total change inentropy

∆S =5

2NkB ln

T3

T1+NkB ln

V3T1

V1T3= NkB ln

T3/23 V3

T3/21 V1

. (5)

(b) The partition function for an ideal monatomic gas is

QN =1

N !QN1 (6)

Problem 1 from Qualifying exam 2008-Fall.

where

Q1 =

∫d3rd3p

h3e−p

2/(2mkBT ) = V

(2πmkBT

h2

)3/2

. (7)

The Helmholtz free energy is then

A = −kBT lnQN = kBTN

(lnN

V

(h2

2πmkBT

)3/2

− 1

)(8)

(c) Given the Helmholtz free energy, the entropy is

S = −(∂A

∂T

)V,N

(9)

= kBN

(lnV

N

(2πmkBT

h2

)3/2

+5

2

). (10)

It follows that the change in entropy is

∆S1→3 = S(N,V3, T3)− S(N,V1, T1) = kBN lnV3T

3/23

V1T3/21

(11)

in agreement with Part (a).

3 Spaceship

A spaceship is sent from earth to another solar system which is at a distance of20 light years from the earth. The spaceship is rapidly accelerated to a velocity(relative to the earth) of 0.99c toward its objective. Upon reaching the distantsolar system, the crew takes some pictures as the ship is rapidly accelerated to0.99c heading back toward the earth. The pictures are immediately transmittedby radio to those waiting on earth. Using special relativity and ignoring therelativistic effects of the periods of acceleration and deceleration:

(a) How long does the round trip take according to people waiting on earth?

(b) How long does the round trip take for people on the spaceship?

(c) After how many years since takeoff do the people on earth see the firstpictures from the distant solar system?

(d) In those pictures, they see the wonders of the distant solar system as wellas the clocks on the spaceship. What time do these clocks read (in years)?Show that this is consistent with the Doppler shifted clock frequencies.

Solution

(a)

t = 40/0.99 = 40.4years (12)

(b)

γ =1√

1− v2/c2= 1/

√0.02 = 1/0.14 (13)

t′ = t/γ = 40.4× 0.14 = 5.7years. (14)

(c)

t = 20/0.99 + 20 = 40.2years (15)

(d) Time=5.7/2 years∼2.8 years.

Doppler shift:

ν =

√1− v/c1 + v/c

ν0 =

√0.01

1.99ν0 ∼ ν0/14 (16)

so time= 40.2years/14 ∼ 2.8years, which is consistent with the above calcula-tion.

Problem 8 from Qualifying exam 2008-Fall.

4 Estimations

(a) The molten metal in a furnace appears to emit predominantly blue light.Is the temperature of the metal closest to (I) 3 K, (II) 3 × 104K, or (III)3 × 106K?

(b) Estimate the ground state energy of a harmonic oscillator using the uncer-tainty principle.

(c) In an electron interferometer, an electron is split into two paths ABC andADC in the figure below and recombined. The rectangular paths havethe dimensions AB = DC =a and AD = BC =L. The whole rectangle isplaced in a uniform electric field along the direction AB or DC. If interfer-ence fringes are produced as the electric field strength is varied, find theinterference fringes’ separation as the change of the electric field in termsof the electron kinetic energy, its mass and charge, the dimensions of thepaths and any universal constants.

Solution

(a) Visible light ∼ 2eV ∼ 2.4× 104K. Therefore, it must be case II.

(b) For the harmonic oscillator: (using ∆p∆x ∼ h)

E =p2

2m+

1

2kx2 ∼ (∆p)2

2m+

1

2k(∆x)2 ∼ h2

2m(∆x)2 +1

2k(∆x)2

In the ground state, find the minimum energy by setting dE/dx = 0:

dEdx∼ −2h2

2m(∆x)3+ k (∆x) = 0→ (∆x)4 = h2

km→ ∆x =

(h2

km

)1/4

→ E = h2

2m(∆x)2+ 1

2k(∆x)2 = h2

2m

(kmh2

)1/2+ 1

2k(h2

km

)1/2

= h2

√km

+ 12h√

km

= hω

Problem 6 from Qualifying exam 2008-Spring.

(c) Along BC:

K = h2k2

2m+ eEa

k =√

(K − eEa) 2mh2

The phase is kL.

Along AD:

K = h2k02

2m

k0 =√

(K) 2mh2

The phase is k0L. So the phase difference ∆φ = (k − k0)L:

(k − k0)L =

(√(K − eEa)

2m

h2 −√K

2m

h2

)L

Rewrite this as:

(k−k0)L =

√K

2m

h2

(√(1− eEa

K)− 1

)L ∼

√K

2m

h2 (1−eEa2K−1)L ∼

√K

2m

h2 (−eEa2K

)L ∼ −eEaL2K

k0

What electric field is needed to shift one fringe? This is the amount needed tochange the phase by 2π:

(k − k0)L ∼ −eEaL2K

k0 = 2π → E =4π

eaL

K

k0=

4π

eaL

√Kh2

2m

5 Thin fiber

A thin fiber of length L is stretched between two supports. The speed ofpropagation of transverse waves on the fiber is c for both polarizations.

(a) What is the contribution of these modes to the heat capacity of the fiberat low temperatures, assuming hc/L kBT? (Don’t worry about solvingthe integral to find the dimensionless prefactor)

(b) What is the heat capacity for hc/L kBT?

Solution

(a) Boson modes: k = nπ/L, (n = 1, 2, 3...). Let ω = kc, then

E = 2∑ hω

ehω/kBT − 1. (17)

For hc/L kBT ,

E ≈ 2L

π

∞∫0

dkhω

ehω/kT − 1=

2L

πc

∞∫0

dωhω

ehω/kT − 1=

2L

πch

(kT

h

)2∞∫

0

dxx

ex − 1

Letting x = hω/kBT and solving for the heat capacity:

C =∂E

∂T∼ 4LkB

2

πhcT

C ∼ T as expected for 1D bosons.

(b) For hc/L kBT , we have:

E = 2∑ hω

ehω/kT − 1= 2

∑ (πhc/L)n

eπhcn/LkT − 1≈ 2

∑ nπhc

Le−πhc/LkT

At low temperatures, only the n = 1 mode is occupied, so we have:

E ≈ 2πhc

Le−πhc/LkT

And the heat capacity is:

C = ∂E∂T

=(πhcL

)2 2kBT

2 e−πhc/LkT

Problem 4 from Qualifying exam 2009-Fall.

6 Estimations 2

For each of the following energy quantities, give an expression in terms offundamental constants which is dimensionally correct (numerical prefactors arenot required) and compute from your expression an order of magnitude estimatein units of electron Volt (eV).Useful information (formulas in Gaussian cgs units):o Rydberg: 1 Ry = mee

4/(2h2) ≈ 13.6 eV.o Bohr Magneton: µB = eh/(2mec) ≈ 6 × 10−5eV/T(T Tesla).o Fine Structure Constant: α = e2/(hc) ≈ 1/137.Example: The electron kinetic energy in the ground state of the Hydrogen atom.Solution: Denote the size of the atom by aB . From Newton’s second law onefinds T = mev

2/2 = e2/(2aB) for the kinetic energy and the quantization con-dition mevaB = h leads to T = h2/(mea

2B). When equated with the potential

energy e2/aB , this yields aB = h2/(mee2) ≈ 5 × 10−11m. The kinetic energy

is then T = mee4/h2 = 2Ry ≈ 10eV. (One Ry is 13.6 eV, but the order of

magnitude 10 eV is sufficient for this problem).

(1) The ground state binding energy of the electron in a U91− ion (a Uraniumion with atomic number 92, atomic weight 238, which is 91-times ionizedso that there is only one electron present).

(2) The ground state energy level splitting in a Hydrogen atom due to thespin-orbit interaction (fine structure splitting).

(3) The hyperfine splitting in a Hydrogen atom (the splitting of electronicenergy levels due to the interaction with the magnetic moment of theproton).

(4) The relativistic correction to the binding energy of a Hydrogen atom.

(5) The Zeeman splitting of electron levels in a Hydrogen atom when themagnetic field is 1 T (T Tesla).

(6) The energy of a nucleon in a typical nucleus.

(7) The rotational energy of an H2 molecule.

Solution

(1) Using the scaling e2 → Ze2 with Z = 92 in the expression for the hydrogen atomwe obtain

E ∼ mZ2e4

h2 ≈ 104Ry ≈ 105eV.

(2) This is the energy of the electron spin dipole in the magnetic field due to itsorbital motion (Er radial electric field, p momentum)

∆E ∼ µBvErc∼ eh

mec

pe

meca2B

∼ e8me

h4c2∼ α2Ry ≈ 10−3eV.

Problem 7 from Qualifying exam 2009-Spring.

(3) It is reduced by a factor of the mass ratio from the fine structure splitting

∆E ∼ me

mpα2Ry ≈ 10−6eV

(4) By Taylor expansion of√m2c4 + p2c2

∆E ∼ mec2(

p2

m2ec2

)2 ∼ α2Ry ≈ 10−3eV.

(5) The Zeeman energy in the magnetic field B is

∆E ∼ µBB ≈ 10−4eV.

(6) In a nucleus of radius rN ∼ 1fm = 10−15m(T kinetic energy)

EN ∼ TN =h2

mNr2N

≈ TNTe

10eV =mer2

N

mNa2B

10eV ≈ 107eV.

(7) The rotational energy is (L angular momentum, µ reduced mass)

Erot ∼L2

µr2∼ h2

mHa2B

∼ me

mHRy ≈ 10−2eV.

7 Photon gas

(a) Consider a photon gas at temperature T = 2.9 K. Using E = uV anddE = −PdV for an adiabatic, quasistatic process, derive an expression forhow the energy density u depends upon the volume V for an adiabatic,quasi-static expansion of the gas. Hint: P = u/3 for a photon gas.

(b) For a photon gas, u ∼ T 4. Using this, derive an expression for how thevolume V depends upon the temperature T for an adiabatic, quasistaticprocess.

(c) If the universe expanded quasi-statically and adiabatically from an initialstate where the radiation temperature was T = 3000 K to the currentstate with a temperature of T = 2.9 K, what is the ratio of the currentvolume to the initial volume? Assume that the universe is composed ofno matter.

Solution

(a)dE = −PdV

E = uV → dE = udV + V du = −PdV = −(u/3)dV

Therefore:−(4/3)udV = V du, or (du/u) = −4/3(dV/V)

This results in:uV 4/3 = constant, or u ∼ V−4/3

(b) Using what we have above, T 4V 4/3 = constant, or V ∼ T−3.

(c) (Vfinal/Vinitial) = (Tinitial/Tfinal)3 = (3000/2.9)3 = 1.1× 109.

Problem 8 from Qualifying exam 2009-Spring.

8 Estimations 3

Estimate each of the following, with an explanation for your reasoning:

(a) the average kinetic energy of a monatomic air molecule in this room.

(b) the molar heat capacity at constant volume of a rock at room temperature.

(c) the electrical resistance of a metal wire one meter long and 1 mm in radius.(Hint: ρ ∼ 10−8Ωm).

(d) the number of atoms in 1 cm2 on the surface of a solid metal.

(e) the number of atoms in 1 m3 of gas at room temperature and atmosphericpressure.

Solution

(a) 3/2kT = 6.2× 10−21J(for T ∼300 K)

(b) CV = 3R = 24.9J/molK

(c) R = ρ(L/A) = 10−8/(π ∗ 10−6) = 0.3× 10−2 ∼ 3× 10−3Ohm

(d) Interatomic spacing of solids is∼ 4×10−10 m. Each atom then occupies an area ona surface of roughly 1.9×10−19m2. For 1 cm2, which has an area of 1×10−4m2,we then have: number of atoms = 1 × 10−4m2/1.6 × 10−19m2 = 6.25 × 1018

atoms

(e) PV = NkT → N = PV/kT = (101×103Pa)(1m3)/(1.38×10−23∗300) = 2.4×1025

Problem 9 from Qualifying exam 2009-Spring.

9 Ideal gas in a cycle 6

One mole of an ideal monatomic gas at an initial volume V1 = 25L and pressureP1 = 105Pa is subjected to the following three step cycle.First, it is heated at constant volume to P2 = 2 × 105Pa.Second, it is then isothermally expanded to V3 = 2V1.Third, the volume is reduced back to V1 at constant pressure.All processes are quasi-static. The gas constant is 8.314J/(mole K).

(a) Draw the P − V diagram of the cycle indicating the pressure and volumeafter each step.

(b) Find the temperature of the gas at each step of the cycle.

(c) Find the heat flow for each part of the cycle.

(d) Calculate the efficiency of the cycle.

Solution

(a)P1 = 105Pa, V1 = 25L, P2 = 2× 105Pa, V2 = 25L, P3 = 105Pa, V3 = 50L.

(b)

T1 =P1V1

R=

105Pa× 25L

8.314 JmolK

× 1mol= 301K

T2 = 2× T1 = 601K

T3 = T2 = 601K

(c)

Q = W + ∆U,∆U = CV ∆T,CV =3

2Rn∆T,W =

∫PdV

1→ 2 : V = const,W = 0,Q12 =3

2R∆T = 3750J

2→ 3 : T = const,∆U = 0,W =

∫PdV = nRT ln

V3

V2= 3466J

Q23 = W = 3466J

Problem 1 from Qualifying exam 2010-Fall.

3→ 1 : P = const,W = P∆V = −2500J

∆U = −3

2Rn ∗ 301K = −3750J

Q31 = W + ∆U = −6250J

(d)

efficiency =W

Qh=

Qh − |Qc|Qh

= 1− Q31

Q12 + Q23= 0.134

10 Thermal neutron

The nucleus113Cd captures a thermal neutron having negligible kinetic energy,producing 114Cd in an excited state. The excited state of 114Cd decays to theground state by emitting a photon. Find the energy of the photon. Usefulconstants:

me = 0.511 MeV/c2 mn = 1.008665 u m(113Cd) = 112.904401 u

m(114Cd) = 113.903359 u 1 u = 931.5 MeV/c2 1 eV = 1.6 × 10−19 J

Solution

The reaction is 113Cd + n→114 Cd + γ.4-vector conservation with c = 1

m113

000

+

mn

000

=

√m2

114 + p2γ

−pγ00

+

pγpγ00

m113 +mn =

√m2

114 + p2γ + pγ

(recovering c)

Eγ = pγ =2m113mn −m2

114 +m2113 +m2

n

2 (mn +m113)· c2

= 9.042Mev.

Problem 3 from Fall02.