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For the alpha particle m= 0.0304 u which gives 28.3 MeV binding energy!

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Page 1: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!
Page 2: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

For the alpha particle m= 0.0304 u which gives 28.3 MeV binding energy!

Page 3: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Is Pu unstable to -decay?23694

Pu U + 23694

23292

42 + Q

Q = (MPu – MU - M)c2

= (236.046071u – 232.037168u – 4.002603u)931.5MeV/u

= 5.87 MeV > 0

Page 4: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!
Page 5: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Beta decay Examine the stability against beta decay by plotting the rest mass energy M of nuclear isobars (same value of A) along a third axis perpendicular to the N/Z plane.

                                                                                   

    

Page 6: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

103.912

103.910

103.908

103.906

103.904

Mas

s, u

42 44 46 48

Atomic Number, Z

A = 104 isobars42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

Odd Z

Even Z

-decay: X Y + AZ N ?

?? N-1A

Z+1

e-capture: X + e Y N N+1A

Z-1 AZ

Page 7: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Fourier Transforms Generalization of ordinary “Fourier expansion” or “Fourier series”

de)(g2

1)t(f ti

de)t(f2

1)(g ti

Note how this pairs canonically conjugate variables and t.

Page 8: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

4/)(

4/)(

220

2

max

EE

E

Breit-Wigner Resonance Curve

Eo E

1.0

0.5

MAX

= FWHM

Page 9: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Incompressible Nucleus

R=roA1/3

ro1.2 fm

Page 10: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

v t

d

Incident mono-energetic beamscattered particles

A

N = number density in beam (particles per unit volume)

N number of scattering centers in targetintercepted by beamspot

Solid angle d representsdetector counting the dN particles per unit time that

scatter through into d

FLUX = # of particles crossing through unit cross section per sec = Nv t A / t A = Nv

Notice: qNv we call current, I, measured in Coulombs.

dN N F d dN = N F d dN = N F d

Page 11: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Nscattered = N F dTOTALThe scattering rate

per unit time

Particles IN (per unit time) = FArea(of beam spot)

Particles scattered OUT (per unit time) = F N TOTAL

AvogadroN

A

N

Cro

ss s

ecti

on

incident particle velocity, v

Page 12: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

D. R. Nygren, J. N. Marx, Physics Today 31 (1978) 46

p d

e

Momentum [GeV/c]

dE

/dx(

keV

/cm

)

Page 13: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Notice the total transition probability t

dE

dNEVE

tP iNtotal

2||

2

and the transition rate

dE

dNEVEtPW iNtotal

2||

2 /

Page 14: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

vx

vy

vzClassically, for free particlesE = ½ mv2 = ½ m(vx

2 + vy2 + vz

2 )

Notice for any fixed E, m this definesa sphere of velocity points all which give the same kinetic energy.

The number of “states” accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere).

dV = 4v2dv

Page 15: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Classically, for free particlesE = ½ mv2 = ½ m(vx

2 + vy2 + vz

2 )

dv mvdE mE

Ed

mv

Eddv

2

We just argued the number of accessible states (the “density of states”) is proportional to 4v2dv

mE

dE

m

ECdvvCdN

2

244 2

dEEm

CdN 2/1

2/32

4

dN

dE E1/2

Page 16: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

fi

fiN

vv

pEVE2

24

2||

Page 17: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

d

parentdaughter

d

parentdaughter

N

NN

tN

tNtN

,0

,0,0

)(

)()(

d

daughtertt

d

parent

d

daughter

N

Ne

tN

tN

tN

tN

,0

,0)( 1)(

)(

)(

)(0

Which we can rewrite as:

y = x m + b

What if there were initially some daughter products already there when the rock was formed?

Page 18: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Rb-Sr dating method

Allows forthe presence of initial 87Sr

Page 19: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Calculation of the kinetic energy of an alpha particle emitted by the nucleus 238U. The model for this calculation is illustrated on the potential energy diagram at right.

Page 20: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

222222II |)(| rkeDrx

In simple 1-dimensional case

E

I II III

V

probability of tunneling to here

x = r1 x = r2

Page 21: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

R

E

Where

2

2

0

)2(2

4

1

r

eZT

r2

So let’s just write

r

eZrV

2

0

)2(2

4

1)(

Tr

rrV 2)(as

Page 22: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

)2()2(

exp|)(| 2 ZRBT

ZAerX

When the result is substituted into the exponential the expression for the transmission becomes

Page 23: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

ml () = Pm

l (cos)eim

Pml (cos) = (1)m(1-cos2m [( )m Pl (cos)]

(2l + 1)( lm)! 4( l + m)!

d d (cos)

d d (cos)Pl (cos) = [( )l (-sin2)l ] 1

2l l!

So under the parity transformation:

P:ml () =m

l (-)=(-1)l(-1)m(-1)m m

l ()

= (-1)l(-1)2m ml () )=(-1)l m

l ()

An atomic state’s parity is determined by its angular momentum

l=0 (s-state) constant parity = +1l=1 (p-state) cos parity = 1l=2 (d-state) (3cos2-1) parity = +1

Spherical harmonics have (-1)l parity.

Page 24: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

In its rest frame, the initial momentum of the parent nuclei is just its

spin: Iinitial = sX

and: Ifinal = sX' + s + ℓ

1p1/2

1p3/2

1s1/2

4He

S = 0 So |sX' – sX| < ℓsX' + sX

Page 25: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

the parity of the emitted particle is (-1)ℓ

mY ~

Since the emitted is described by a wavefunction:

Which defines a selection rule: restricting us to conservation of angular momentum and parity.

If P X' = P X then ℓ = even

If P X' = P X then ℓ = odd

Page 26: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

eeee

eNif ppch

dppTT

gW )()(

6437

22

max2

2

4

PM

This does not take into account the effect of the nucleus’ electric charge which accelerates the positrons and decelerates the electrons.

Adding the Fermi function F(Z,pe) , a special factor (generally in powers of Z and pe),

is introduced to account for this.

37

22

max224 ))(,(64)(

ch

dppTTpZFgdpp ee

eeNee MP

Page 27: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

This phase space factor determines the decay electron momentum spectrum. (shown below with the kinetic energy spectrum for the nuclide).

Page 28: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

the shortest half-lifes (most common) -decays “super-allowed”

0+ 0+

10C 10B*14O 14N*

The space parts of the initial and final wavefunctions are idenitical!

What differs?

The iso-spin space part (Chapter 11 and 18)

|MN|2 = 2

Page 29: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Note:

the nuclear matrix element depends on how alike A,Z and A,Z±1 are.

When A,Z A,Z±1 |MN|2~ 1

otherwise |MN|2 < 1.

If the wavefunctions correspond to states of different J or different parities

then |MN|2 = 0.

Thus the Fermi selection rules for beta decay J = 0 and

'the nuclear parity must not change'.

Page 30: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Total S = 0 (anti-parallel spins) Total S = 1 parallel spins)

Fermi Decays Gamow-Teller Decays

Nuclear I = 0 Ii = If + 1

I = 0 or 1

With Pe, = (1)ℓ = +1

PA,Z = PA,Z1

I = 0,1 with no P change

Page 31: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

10C10B*

14O14N*0+ 0+ Fermi Decays

6He6Li

13B13C n p 3H3He 13N13C

0+ 1+

3/2 1/2 Gamow-Teller Decays

e, pair account for I = 1 changecarried off by their parallel spins

1/2+ 1/2

1/2+ 1/2 1/2 1/2

Page 32: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Forbidden Decays

ℓ=1 “first forbidden” With either Fermi decays s = 0Gamow-Teller decays s = 1

2 ,1 ,0I

with Parity change!

)22(

)01(

)2/52/1(

*122122

7676

1717

SnSb

SeBr

ON

Page 33: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Forbidden Decays

ℓ=2 “second forbidden” With either Fermi decays s = 0Gamow-Teller decays s = 1

2IWith no Parity change!

)2/32/7(

)03( 137137

2222

BaCs

NeNa

even rarer!

Fermi and Gamow-Teller already allow (account for) I= 0, 1 with no parity change

Page 34: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

If this change is large enough, the will not be absorbed by an identical nucleus. 2

2

2 cm

pQE

N

emitted

In fact, for absorption, actually need to exceed the step between energylevels by enough to provide the nucleus with the needed recoil:

p=E/cTN =

pN2

2mN

= p2

2mN

The photon energy is mismatched by

2

2

2 cm

EQE

N

absorbed

2

2

2

2

22

cm

E

cm

E

NN

Mössbauer Effect

Page 35: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

57Co7/2

5/2

EC

=270d

57Fe1/23/2 14.4keV

136keV

=10-7s

As an example consider the distinctive 14.4 keV from 57Fe.

The recoil energy of the iron-57 nucleus is

this is 5 orders of magnitude greater than the natural linewidth of the iron transition which produced the photon!

eVGeV

keV

cm

EE

N

recoil

002.0)022.53(2

)4.14(

22

2

2

With = 107 s, =108 eV

~90% of the 57Fe* decays are through this intermediate level produce 14.4 keV s.

Page 36: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

BB

BeC

LiN

O

HeO

pF

10

5

10

5

8

4

12

6

6

3

14

7

16

8

3

2

17

8

19

9

BB

BB

BeC

BeC

LiN

LiN

O

HeO

HF

dF

Ne

nNe

pF

11

5

9

5

10

5

10

5

9

4

11

6

8

4

12

6

7

3

13

7

6

3

14

7

16

8

3

2

17

8

3

1

17

9

18

9

20

10

19

10

19

9

2010[ Ne]*

The Compound Nucleus

Page 37: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

To quantum mechanically describe a particle being absorbed, we resort to the use of a complex potential in what is called the optical model.

Consider a traveling wave moving in a potential V then this plane wavefunction is written

The Optical Model

ikxeψ 22 /)(8 VEmk

If the potential V is replaced by V + iW then k also becomes complex and the wavefunction can be written

where

xikxik eeψ 21 and now here21

ikkk

Page 38: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

PdU 119

46

238

922

A possible (and observed) spontaneous fission reaction

238U119Pd

8.5 MeV/A

7.5 MeV/A

Gains ~1 MeV per nucleon!2119 MeV = 238 MeV

released by splitting

Page 39: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

The potential energy V(r) = constant-B as a function of the separation, r, between fragments.

Z2/A=36

Z2/A=49

such unstable statesdecay in characteristicnuclear times ~10-22 sec

Tunneling does allow spontaneous fission, but it must compete with

other decay mechanisms (-decay)

Page 40: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

At smaller values of x, fission by barrier penetration can occur, However recall that the transmission factor (e.g., for -decay) is

eXwhere

drh

ErVm ])([22 m

while for particles (m~4u)this gave reasonable, observable probabilities for tunneling/decay

for the masses of the nuclear fragments we’re talking about, can become huge and X negligible.

Page 41: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

Natural uranium (0.7% 235U, 99.3% 238U)

undergoes thermal fission

onlythe

Fission produces mostly fast neutrons

Mev

but is most efficiently induced by slow neutrons

E (eV)

Page 42: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

1

3

20

1

11

2

1 HeHH Q=5.49 MeV

eHHH1

2

10

1

10

1

1 Q=0.42 MeV

The sun 1st makes deuterium through the weak (slow) process:

)(2 0

1

12

4

21

3

21

3

2 HHeHH Q=12.86 MeV

then

2 passes through both of the above steps then can allow

This last step won’t happen until the first two steps have built up sufficient quantities of tritium that the last step even becomes possible.

The proton-proton cycle

2(Q1+Q2)+Q3=24.68 MeVplus two positrons whose

annihilation brings an extra

4mec 2 = 40.511 MeV

Page 43: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

eCN 7

13

66

13

7Q=1.20 MeV

6

13

70

1

16

12

6 NHC Q=1.95 MeV

7

14

70

1

17

13

6 NHC Q=7.55 MeV

The CNO cycle

7

15

80

1

17

14

7 OHN Q=7.34 MeV

eNO 8

15

77

15

8Q=1.68 MeV

2

4

26

12

60

1

18

15

7 HeCHN Q=4.96 MeV

carbon, nitrogen and oxygen are only catalysts

Page 44: For the alpha particle  m= 0.0304 u which gives 28.3 MeV binding energy!

The 1st generation of stars (following the big bang) have no C or N.The only route for hydrogen burning was through the p-p chain.

Shown are curvesfor solar densities

105 kg m-3 for protons and 103 kg m-3 for 12C.

Rat

e of

en

ergy

pro

du

ctio

n

In later generationsthe relative importance of the two processes depends upon temperature.