FSU f2010 Exam Solutions

8/3/2019 FSU f2010 Exam Solutions

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Physics Qualifying Examination

Problems 16 Thursday, August 26, 2009 15 pmProblems 712 Friday, August 27, 2009 1-5 pm

1. Solve each problem.

2. Start each problem solution on a fresh page. You may use multiple pages per

problem.

3. At the top of each solution page put the problem number (112) and your

Social Security number, but not your name or any other information.

Academic Honor Code

Students are expected to uphold the Academic Honor Code published in The Florida

State University Bulletin and the Student Handbook. The first paragraph is:

The Academic Honor System of Florida State University is based on the premise that

each student has the responsibility (1) to uphold the highest standards of academic

integrity in the student's own work, (2) to refuse to tolerate violations of academic

integrity in the University community, and (3) to foster a high sense of integrity and

social responsibility on the part of the University community.

Specific rules for this exam:1. Sit only in your assigned seat.2. All cell phones must be turned off during the exam.3. You are allowed to bring a hand calculator and a book of math tables, but not one

that has a list of physics formulas (e.g. Maxwell's equations or the equations of

fluid flow or thermodynamics, etc) or physical constants (e.g. electron mass,

acceleration of gravity, etc). If such information should be supplied, it will beincluded in the statement of the problem (and that is generally the case with

numbers like the electron mass, Planck's constant, etc.).

4. you may ask questions to the faculty proctor during the exam. If it is decidedsomething is unfairly left out, that will be announced to everyone in the room.

5. You must not receive any information verbal, written, or otherwise fromanyone during the exam except the faculty proctor.

6. You must not give any information to anyone during the exam.


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Problem 1

One mole of an ideal monatomic gas at an initial volume V1 = 25L and pressure

P1 = 105Pa is subjected to the following three step cycle.

First, it is heated at constant volume toP2 = 210

5

Pa.

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 8.314 J/(mole K).

(a)Draw thePVdiagram of the cycle indicating the pressure and volume after eachstep.

(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.


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Problem 1


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Problem 2

A mass m is hung from a fixed support on the end of a light unstretchable cord of length

L forming a simple pendulum. The mass is pulled to the side so that the cord makes an

angle of1 with the vertical.

(a) Find the potential energy of the system relative to its lowest position.

(b) If the mass is released at1 from rest, determine the velocity of the mass when it is at

its lowest position.

(c) Find both the potential and kinetic energies of the mass when its cord makes an angle

2 < 1 with the vertical if the mass is released at 1 from rest.

Solution


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Problem 3

Let the Lagrangian of two interacting point particles be

2 21 21 2

1 22 2

m m L r r

r r

= +

1. Explain why the energy of this system is, or is not, conserved.2. Write down the expression for the center of mass ( )R t of the system. How is the

center of mass reference frame defined?

3. Let and express the Lagrangian in the center of mass reference frame asa function of the coordinate

1r r r=

2

r

and the velocity r .

4. Rewrite the resulting Lagrangian in spherical coordinates (polar angle ,azimuthal angle , and magnitude r r

).=

5. Write down the Euler-Lagrange equations for the two angles and identify fromone of these equations a conserved quantity.

6. Prove that the motion is confined to a plane.


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Problem 3 Solution


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Problem 4

Consider two donut shaped permanent magnets with magnetization parallel to thez-axis(vertical in the figure), which slide without friction on a vertical rod. Treat the magnets asdipoles, with mass Mand dipole momentm. If you put the magnets back-to-back (NS-SN), the upper magnet will float.

a)

Determine the height at which it floats.b) Determine the angular frequency of small oscillations about the equilibriumheight.

c) Use dimensional analysis to determine the frequency dependence of the powerradiated by the oscillating magnetic dipole.


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Solution:

a) The interaction energy between the two anti-aligned dipoles is3

2

0

2 z

mUm

=

So including the gravitational energy we have Mgzz

m

Utot += 32

0

2

The equilibrium height is determined by the minimum of the total energy with respecttozso

Mgz

mU

z eqtot +=

=

4

2

0

2

30

41

2

0

2

3

=

Mg

mzeq

b) From the Lagrange equations of motion we have Mgzm

zM = 42

02

3

Expanding near the equilibrium height we haveeqz

zgz

4=

From here we see that

21

4

=

eqz

g

c) The power radiated must have units of energy/time. The energy must be thedipole energy and so dimensionally

3

2

Length

mP

The length-scale is obtained from the speed of light and frequency so4P


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Problem 5

Consider two identicalspin particles with mass m which move in 3D and attract each

other via harmonic potential with spring constant .

a) Write down the Hamiltonian for this system.b) Determine the energy spectrum.c) Assuming a center of mass momentum coordinate system, determine the spin

structure of the lowest and the first excited state(s). What are the degeneracies ofeach of these states?

Solution:

a) 2212

2

2

1 )(2

1

22rr

m

p

m

pH

++=

b)

++=

2

122n

m

KE

where n=0,1,2 andmmred

2==

c) The ground state of a harmonic oscillator is even under parity transformation andtherefore even under exchange. By Pauli exclusion principle, this means that the

spin wavefunction is odd and therefore it is a singlet. This level is non-degenerate.

There are three first excited states of the harmonic oscillator and each is odd

under parity and therefore odd under exchange. Therefore, each is spin triplet.There are therefore 9 first excited states.


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Problem 6

A water molecule can vibrate in the flexing mode in which the hydrogen atoms movetowards and away from each other without stretching the HO bond. The oscillation of this

mode is approximately harmonic with a frequency of 5.0 x 1013 Hz. What is the

probability of a water molecule in its flexing ground state and in the first excited state?Assume that the water is in equilibrium at room temperature. The Boltzmann constant kis 1.38 x 10-23 J/K and Plancks constant h is 6.63 x 10 -34 J.s.

Solution:

The energy levels of a harmonic oscillator are given by En = (n +1/2) hf, where f is the

frequency and n = 0, 1, 2, 3,------- .

So the partition function isZ = e-(1/2)hf/kT + e-(3/2)hf/kT + e-(5/2)hf/kT + e-(7/2)hf/kT + --------------

At T= 300 K, hf/kT = 8.01.

Therefore,Z= e-3.85 + e-11.5 + e-19.2 + e-27.0+ --------- = 0.0187

The probability of being in first excited state is P1 = e -(3/2)hf/kT/ Z = e-11.5/0.0215 = 3.3 x 10-4.

The probability in being ground state is e-(3/2)hf/kT/ Z = e-3.85/0.0215 = .9997.


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Physics Qualifying Examination

Problems 16 Thursday, August 26, 2009 15 pm

Problems 712 Friday, August 27, 2009 1-5 pm

1. Solve each problem.

2. Start each problem solution on a fresh page. You may use multiple pages perproblem.

3. At the top of each solution page put the problem number (112) and yourSocial Security number, but not your name or any other information.

Academic Honor Code

Students are expected to uphold the Academic Honor Code published in The Florida

State University Bulletin and the Student Handbook. The first paragraph is:

The Academic Honor System of Florida State University is based on the premise that

each student has the responsibility (1) to uphold the highest standards of academic

integrity in the student's own work, (2) to refuse to tolerate violations of academic

integrity in the University community, and (3) to foster a high sense of integrity and

social responsibility on the part of the University community.

Specific rules for this exam:

7. Sit only in your assigned seat.8. All cell phones must be turned off during the exam.9. You are allowed to bring a hand calculator and a book of math tables, but not one

that has a list of physics formulas (e.g. Maxwell's equations or the equations of

fluid flow or thermodynamics, etc) or physical constants (e.g. electron mass,acceleration of gravity, etc). If such information should be supplied, it will be

included in the statement of the problem (and that is generally the case with

numbers like the electron mass, Planck's constant, etc.).

10.you may ask questions to the faculty proctor during the exam. If it is decidedsomething is unfairly left out, that will be announced to everyone in the room.

11.You must not receive any information verbal, written, or otherwise fromanyone during the exam except the faculty proctor.

12.You must not give any information to anyone during the exam.


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Problem 7

A non-relativistic particle moves in one dimension along thex-axis with momentum p in

the positive direction. Atx = 0 there is a very narrow potential wall described by potential

2

2( ) ( )

8hV x x

m

=

whereis a positive parameter to characterize strength of the wall and (x) is the Diracdelta function.

(a) Calculate the reflection and transmission coefficients.

(b) Calculate the scattering states and the bound state for negative .


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Solution


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Problem 8


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Problem 9


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Problem 10

Neutron scattering is often done by cooling fast neutrons, which are among the fission

products from a nuclear reactor, by thermalizing them in a moderator such as solid

deuterium oxide (D2O) ice) held at constant cryogenic temperature T. The averagekinetic energy of neutrons from such a source is

21 3

2 2m v kT = = where m is the neutron mass,

2v is the mean-square velocity, and

kis Boltzmanns constant. The neutron mass is mn = 1.68 x 10-27 kg.

a) What is the de Broglie wavelength n for neutrons, in terms of their energy, mass,and Plancks constant?

b) What is the de Broglie wavelength for neutrons with the average kinetic energy interms of the temperature of the source?

c) Hydrogen freezes at 20.25K, and can be used to keep the D2O moderator at afixed temperature. What is the de Broglie wavelength n (in ) for the neutrons

with the average energy corresponding to that temperature? What is the average

neutron energy E (in eV)?

d) The neutrons are monochromatized at the 20.25K thermal energy maximum nand diffracted off a crystalline target with a lattice spacing of 4.21 . What is the

scattering angle that would give Bragg diffraction of neutrons from this crystal?Use the convention for diffraction shown in the figure.


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Solution

Neutron scattering is often done by cooling fast neutrons, which are among the fissionproducts from a nuclear reactor, by thermalizing them in a moderator such as solid

deuterium oxide (D2O) ice) held at constant cryogenic temperature T. The average

kinetic energy of neutrons from such a source is21 3

2 2m v kT = = where m is the neutron mass,

2v is the mean-square velocity, and

kis Boltzmanns constant. The neutron mass is mn = 1.68 x 10-27

kg.

a) What is the de Broglie wavelength n for neutrons, in terms of their energy, mass,and Plancks constant?

The de Broglie wavelength is = h/p= h/(mv), and since E= mv2/2

(note that we are assuming v


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d) The neutrons are monochromatized at the 20.25K thermal energy maximum nand diffracted off a crystalline target with a lattice spacing of 4.21 . What is the

scattering angle that would give Bragg diffraction of neutrons from this crystal?

Use the convention for diffraction shown in the figure.

Referring to the diagram, we see that the path length difference forneutrons which scatter offadjacentplanes in the crystal is just2dsin(), sothat we get constructive interference when this is an integer number of deBroglie wavelengths, i.e.:

n =2dsin()so that we have

5.587sin( ) 0.6635

2 2(4.21 )

nn n

d

= = =

( )

and we see that n=1 is the only possibility for .

Giving an angle of

1sin 0.6635 41.56 = =


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Problem 11

A superconducting (SC) solenoid is to be wound from superconducting wire having a

critical current density 510 /c2 J A cm= at 15 T and a diameter of 2 mm ( is the

current density at which the resistance of the wire suddenly becomes very large compared

to the resistance of the copper). The wire is clad with a copper sheath of thickness 0.5mm. The magnet has a field of 15 T, a bore (inner diameter) of 1 m, and is 2 m long.Remember that the resistance of a superconductor in the superconducting state is exactly

zero.

cJ

Use the long solenoid approximation to determine the characteristics requested below,

including the approximation that the flux is confined to the innermost winding. You may

also assume the layer thickness is very small compared with the inner diameter of the

solenoid, obviously a good approximation. (Note this magnet is approximately the size ofthe SC outsert of the new hybrid under construction at the NHMFL. These are to be only

back of the envelope calculations to give some idea of the parameters).

7

0 4 10 / m T m= :A the resistivity of Cu is 81.7 10 m .

a) What is the maximum current the SC wire can carry and remain superconductingat 15 T?

b) How many layers of wire are needed to produce the 15 T field, and how manytotal turns?

c) What is the resistance of the copper sheath?d) What is the inductance of the solenoid?e) How much energy is stored in the solenoid when it is at full field and

superconducting?

For the parts below assume the Cu part of the coils is shorted so the circuit is a resistance

and inductance in series.

f) If the magnet quenches, that is if the resistance of the SC suddenly becomes verylarge, what will be the maximum voltage across the solenoid (and thus the Cusheath)?

g) Find the current in the Cu sheath as a function of the time after quench. Assumethe equivalent circuit is just the Cu sheath of the solenoid shorted with a

conductor of negligible resistance.

See the figure on the next page.


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Solution:


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Problem 12

Monochromatic blue light with a wavelength of 434.2 nm is incident on a

sample of cesium. Electrons emitted from the cesium surface are observed

having velocities ranging up to 5.491 x 105

m/s. Note that me = 0.511

MeV/c2, qe = 1.602 x 10-19 C, hc = 1240 eV.nm.

e) What is the work function for this sample of cesium?f) Explain why there is arange of emitted electron energies.g) What is the wavelength of the fastest emitted electrons?h) Now assume that the hydrogen discharge lamp produces 2.0 W of

power radiated in this particular blue Balmer series spectral line. Ifthe lamp can be considered to be a point source and emits light

isotropically, estimate how many blue photons per second strike a

circular cesium sample 7.5 cm in diameter and placed 100 cm in front

of the lamp.


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Solution:

Monochromatic blue light with a wavelength of 434.2 nm is incident on a

sample of cesium. Electrons emitted from the cesium surface are observed

having velocities ranging up to 5.491 x 105

m/s. Note that me = 0.511MeV/c2, qe = 1.602 x 10-19 C, hc = 1240 eV . nm.

a) What is the work function for this sample of cesium?The photoelectric effect defines the kinetic energy of emitted

electrons

E = + KEmaxthus: = E - KEmax

= hc/1/2 mev

2max

= (1240 eV.nm)/(434.2 nm)

1/2 [0.511 MeV/(3x10

8m)

2]( 5.491 x 10

5m/s)

2max

= 2.856 eV 0.856eV = 2.00eV

b) Explain why there is arange of emitted electron energies.The valence electrons within a metal form a continuum of energies up

to an energy below the ionization threshold. When a photon strikes

the surface, the electrons absorb the full photon energy. Those

electrons nearest the surface can leave the surface with the maximum

energy available, which is the photon energy minus the work function,

. Electrons deeper inside the metal are more likely to interact with

electrons around other atoms on their way out of the sample before

leaving the surface, and these interactions will reduce their kinetic

energies.

c) What is the wavelength of the fastest emitted electrons?The deBroglie wavelength of electrons is

e e

h h hc

p m v m vc= = =

= (1240 eV.nm)/ {[0.511 MeV/(3x10

8m)][5.491x10

5m/s]}

.= 1 326 nm

d) Now assume that the hydrogen discharge lamp produces 2.0 W ofpower radiated in this particular blue Balmer series spectral line. If

the lamp can be considered to be a point source and emits light

isotropically, estimate how many blue photons per second strike a


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circular cesium sample 7.5 cm in diameter and placed 100 cm in front

of the lamp.

The energy density per time (from an isotropic source) is given by the

inverse square law of light, and energy per time is power;

R

24PIR=

E/t=P

The fraction of that intensity which reaches the cesium sample has

the same ratio as the solid angle subtended by that sample has to the

solid angle of a full sphere (4).

4

EI t

EIt

= =

(1)

The fraction of that intensity which reaches the cesium sample has

the same ratio as the solid angle subtended by that sample has to the

solid angle of a full sphere (4).It is actually a rather difficult geometrical problem to find the area

cut out of a sphere by a circular disk which has its circumference

touching the inside of the sphere. This fraction would give us the solid

angle subtended by the cesium sample, and therefore also give us the

fraction of the total energy absorbed by the sample. To attack this

problem we will look at a cross-sectional cut of the disk and1

sr

2

sphere.

We can visualize the disk edge-on as the solidline from 1 to 2, contained inside the sphere thus:

If the radius of the disk, r, is small compared to the radius of the

sphere,R, then we can use the small angle approximation; that is:

tan (=s/R) sin (= r/R)which gives that r s.

If this is true, the solid angle subtended by the disk ( = s2/R2)can be approximated by the surface area of the disk divided by the

radius of the sphere squared ( = r2/R2).The fraction of the intensity reaching cesium sample can thus be

calculated from (1) above:

[ ] [ ]4 4

I I E t E t

= =


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[ ] [ ]2 2 2

24 4

r R r E t E t P

R

= =

[ ]

2 2

2 24 4

r r E Pt Pt R R

= =

[ ]( )

( )

which can be solved:

( )2

2

7.5 / 20.0003516

4 100

cm E Pt Pt

cm

= =

( )( )( )60.0003516 2 10 1 x watts sec=10

9197.03 10 4.39 101.602 10 /

x J x eV x J eV

= =

This is total energy deposited per second. Because this comes in

discrete packets of energy (photons), which each have an energy

E= h= hc/, the number of photons can be calculatedEtotal = N

.E

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FSU f2010 Exam Solutions