For the alpha particle m= 0.0304 u which gives 28.3 MeV binding energy!
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Transcript of 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!
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
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.
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
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.
4/)(
4/)(
220
2
max
EE
E
Breit-Wigner Resonance Curve
Eo E
1.0
0.5
MAX
= FWHM
Incompressible Nucleus
R=roA1/3
ro1.2 fm
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
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
D. R. Nygren, J. N. Marx, Physics Today 31 (1978) 46
p d
e
Momentum [GeV/c]
dE
/dx(
keV
/cm
)
Notice the total transition probability t
dE
dNEVE
tP iNtotal
2||
2
and the transition rate
dE
dNEVEtPW iNtotal
2||
2 /
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
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
fi
fiN
vv
pEVE2
24
2||
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?
Rb-Sr dating method
Allows forthe presence of initial 87Sr
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.
222222II |)(| rkeDrx
In simple 1-dimensional case
E
I II III
V
probability of tunneling to here
x = r1 x = r2
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
)2()2(
exp|)(| 2 ZRBT
ZAerX
When the result is substituted into the exponential the expression for the transmission becomes
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.
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
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
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
This phase space factor determines the decay electron momentum spectrum. (shown below with the kinetic energy spectrum for the nuclide).
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
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'.
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
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
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
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
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
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.
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
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
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
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)
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.
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)
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
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
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.