Transcript chapter 8

Chapter 8
Alpha Decay
◎ Introduction and some other properties of α-decay
● The simple theory of Coulomb barrier penetration
◎
The angular momentum barrier
§ 8-1 Introduction and some other properties of α-decay
1. The theory of α-decay is an application of simple quantum
mechanics and its presentation will not, in the simple approach that
we shall adopt, add much to our knowledge of nuclear structure.
2. We should answer the following question: why do the mean lives of
α -emitting nuclei vary so dramatically
Th 228
88 Ra  
from τ= 2.03 × 1010 years for
232
90
to τ= 4.3 × 10-7 seconds for
212
84
Po208
82 Pb  
This is a 24 orders of magnitude
variation in transition rates !!
3. There are many unstable heavy nuclei in nature. They tend to give
away excessive energies and charges by emitting α- particles. Only a
few will undergo nucleon emission.
4. We need to find out a reasonable explanation of why the α- particle
decay turns out to be the preferable choice for a decay process to
occur in unstable heavy nuclei.
5. An α- particle being kicked out from an unstable nucleus is
basically the effect of Coulomb repulsion. An α- particle is much
less massive than a parent nucleus and has a more stable structure
with large binding energy (EB = 28.3 MeV). A heavy nucleus with
too many protons can reduce some Coulomb repulsion energy by
emitting an α- particle.
Energy Release (Q value) for Various Modes of Decay of 232U
Emitted
Particle
Energy
Release (MeV)
Emitted
Particle
Energy
Release (MeV)
n
-7.26
4He
+5.41
1H
-6.12
5He
-2.59
2H
-10.70
6He
-6.19
3H
-10.24
6Li
-3.79
3He
-9.92
7Li
-1.94
α-particle decay
( Z , A)  ( Z  2, A  4) 42 He
Q  [ M ( Z , A)  M ( Z  2, A  4)  M (2,4)]c 2
The available energy Qα
goes into the kinetic
energies of the α-particle
and of the recoil of the
daughter nucleus.
The energy-level diagram for two
nucleic connected by α-decay
The energy-level diagram for the
α-decay of 242Pu
If Qα > 0, α-decay is
energetically possible;
however, it may not
occur for other reasons.
We now have to apply the energy conditions for α-decay to occur in real nuclei
and to find where in the periodic table it is expected to occur.
( Z , A)  ( Z  2, A  4) 42 He
Rewrite the definition of Qα in terms of the nuclear binding energies.
Q  B(Z  2, A  4)  B(2,4)  B( Z , A)
dB d  B 

A 
dA dA  A 
d  B  B  dA 
 A    
d A  A  A  dA 
d B B
 A  
dA  A  A
(1)
Thus α-decay is energetically allowed if
B (2,4)  B ( Z , A)  B ( Z  2, A  4)
dB
 d ( B / A) B 
4
 4 A
 
dA
dA
A

(2)
B (2,4)  B ( Z , A)  B ( Z  2, A  4)
4
Above A ≈ 120, d(B/A)/dA is
about −7.7×10−3 MeV. Now
B(2,4), the helium nuclear
binding energy, is 28.3 MeV,
so the critical A must satisfy
the following relation:
A = 151
B

28.3  4  7.7  10 3 A
A

dB
 d ( B / A) B 
 4 A
 
dA
d
A
A

which is
B
 7.075  7.7  10 3 A
A
(3)
Above this A the inequality of equation (3) is satisfied by most nuclei and α-decay
becomes, in principle, energetically possible. In fact from A = 144 to A = 206, 7 αemitters are known amongst the naturally occurring nuclides.
From A = 144 to A =206
From A = 144 to A =206, there are 7 α-emitters of naturally occurring nuclides.
When α-emitters are found in this range of A, the energies of the emitted α-particle
are normally less than 3 MeV. It is known that the lower the energy release the
greater is the lifetime. Their existence implies mean lifetimes comparable to or
greater than the age of the earth (about 4 × 109 years). Most nuclei in this range on
the line stability may be energetically able to decay by α-emission. They do not do so
at a detectable level because the transition rate is too small.
Above Z = 82 (A > 206)
Above Z = 82 many naturally occurring α-emitters are found, many with short lives.
Why are they to be found when their lifetime is so short?
Most of the heavy nuclei to be found on earth were
probably produced in one or more supernova explosions
of early massive stars. Such explosions can produce very
heavy nuclei including trans-uranic elements (Z > 92)
and their subsequent decay by α-emission will take them
down the periodic table in steps of ΔA = −4. Each α-decay
increases the ration N/Z until a β- decay intervenes to
restore the nucleus closer to the line of stability.
7 α-emitters of naturally occurring nuclides.
α- emitter
Natural Abundance
Mean life τ
23.8%
1.04×1016 years
147Sm
125
15.1%
2.74×1011 years
190Pt
112
0.0127%
8.51×1011 years
192Pt
114
0.78%
≈ 1015 years
209Bi
126
100%
3×1017 years
232Th
142
100%
2×1010 years
146
99.2739%
6.3×109 years
144Nd
238U
84
The age of the earth is ~ 4×109 years.
Very long lifetime
comparable to the
age of the earth
Relatively
long lifetime
Fast-decaying
daughter nuclei are in
secular equilibrium.
Early observations on α-decay established that,
for a unique source, the majority of the emitted αparticles had the same kinetic energy.
For each α-emitter, this kinetic energy, Tα, is a
fraction MD/(MD+Mα) of Qα where MD and Mα are
the masses of the daughter nucleus and of an αparticle respectively.
From the previous transparency we see the values
of Qα and the mean life of the principal α-emitter
in one of the naturally occurring radioactive
series.
It is clear that that transition rates (ω) are
a strong function of the kinetic energy.
R α is the range in air at 15ºC
and 1 atmosphere pressure of
the α-particles emitted in a
decay with transition rate ω.
The empirical rule connection the two is known as
the Geiger-Nuttal rule (1911).
log 10   B log 10 R  C
(4)
Z = 90, A ≈ 236
neutron
proton
About 6 MeV which
is the nucleon
separation energy
Effective potential
for an α-particle
6 MeV
For heavy nuclei, the nucleon
separation energy is about 6 MeV, so
the nucleons fill energy levels up to
about 6 MeV below zero total energy.
If two protons and two neutrons from
the top of the filled levels amalgamate
into an α-particle, the binding energy
of 28.3 MeV is sufficient to provide the
four separation energies and leave the
α-particle with positive energy of about
4 MeV.
Now we have an α-particle with positive energy leaving it in a potential well.
The effective potential is the result of nuclear and Coulomb repulsion potentials.
An α-particle is able to tunnel through the “Coulomb barrier” and become free.
The tunneling probability can be calculated quantum mechanically.
This is the effective mechanical potential for an α-particle as
a function of distance between the center of the α-particle and
the center of the system which is the parent nucleus less the
α-particle.
The whole range of potential is separated into three regions:
Region I At distances less than R, approximately the
nuclear radius, the α-particle is in a potential
well of unspecified depth but representing
the effect of the nuclear binding force on the
α-particle.
Region II At a distances R this potential becomes
positive and reaches a maximum value of
U(R) = zZe2/4πε0R, where z = 2 and Z is
the atomic number of the remaining nucleus.
Region III At a distances greater than R the potential
is Coulomb, U(r) = zZe2/4πε0r.
If the parent nucleus Z+2, is energetically capable of
emitting an α-particle of kinetic energy T α, then there
are two possibilities:
(1). T α > U(R): the α-particle, if inside the nucleus, is
free to leave and will do so almost instantaneously.
(That means in a time comparable to the time taken
for the α-particle to cross the nucleus, which is
less than 10-21 second.)
We need to find the barrier
penetration probability.
(2). T α < U(R): classically the α-particle is
confined to the nucleus. Quantum mechanically
it is free to tunnel through the potential barrier,
emerging with zero kinetic energy at radius b
(where b = zZe2/4 πε 0 T α, z = 2) and to move
to large r, where it will have the full kinetic
energy T α.
First we consider a simple square
potential barrier with height U and
thickness t. The whole area can be
separated into three parts.
0, -   r  0

V (r )   U ,
0r t
 0,
r t

The wave function of the α-particle must satisfy the Schrödinger’s equation.
 2 d 2u

 V (r )u  Eu
2
2m dr
(6)
m is the mass of the α-particle
u is the wave function of the α-particle [u = u (r)]
E is the energy of the α-particle
(5)
Solutions of the Schrödinger’s equation in three different sections:
(I)
(7)
(II)
u1  eikr  Be  ir ,
u2  e Kr  e  Kr ,
(III) u3  Ceikr  De  ir ,
k
2mT
p



K
k
2m(U  T )

2mT
p



where T is the kinetic energy
of the α-particle and p is its
linear momentum.
In the section (III) there is no
reflection wave therefore D = 0.
The probability of transmission
is then proportional to |C|2.
r=0
r
r=t
Since the wave function u(r) and its first derivative du/dr are continuous on boundaries
we are able to summarize the following equations:
(1)
On the boundary r = t,
We may have

u2 (t )  u3 (t )
C  ik  ( ik  K ) t
1   e
2 K
du2
dr


r t
du3
dr r t
C  ik  (ik  K ) t
1   e
2 K
(8)
r=0
(2)
On the boundary r = 0,
We may have
u2 (0)  u3 (0)
1 B    
r
r=t
du2
dr

r 0
1 B 
du3
dr
r 0
K
(   )
ik
(9)
Combine four equations from (8) and (9) we have the following relation:
e ikt 
C  ik  K   Kt  ik  K  Kt 
1  1   e  1  1   e 
4  K  ik 
 K  ik  
In evaluating the quantity K
K
(10)
2m(U  T )

U  60 MeV, T  5 MeV  K ~ 2.3 fm -1
But t ~ several fm
The value of e-Kt is extremely small and can be neglected.
e  ikt 
C  ik 
K  Kt
1

1



e
4  K  ik 
(11)
(13)
Tα (MeV)
lnTα
3.6999 - 4.6791(Tα)-1/2
4.0
1.38629
1.36035
4.5
1.50408
1.49415
5.0
1.60948
1.60734
5.5
1.70475
1.70473
6.0
1.79176
1.78967
6.5
1.87180
1.86461
7.0
1.94591
1.93137
(19)
This is a plot of lnω against the values of
Z(M/MαQα)1/2 for the ground state to ground
state transitions of many of the naturally
occurring and man-made α-active elements.
There are deviations from a single straight line
but there is a general tendency for the points to
cluster near a linear relation between lnω and
Z(M/MαQα)1/2 with a slope somewhat less
steep than ̶ 3.97 MeV1/2. The deviation is due
to the neglect of the term f’ of equation (19).
Thus we have a theory which goes some way
towards adequately explaining the range of
these mean lives.
§ 8-3 The angular momentum barrier
There is an important effect in all processes involving nuclear and particle reactions
and decays. We want to introduce the idea in the context of α-particle decay.
Consider the α-particle decay:
( Z  2, A  4)  ( Z , A) 42 He
Suppose the parent and daughter nuclei have spins of quantum number jp and jD.
The total angular momentum must be conserved. If jP≠jD the α-particle must
emerge with relative orbital angular momentum (with quantum number l) with
respect to the recoiling daughter nucleus.
With the zero spin of the α-particle the conservation of the vector of angular
momentum requires that:
jD  jP  l  jD  jP
(20)
jD  jP  l  jD  jP
(20)
The quantum number l must be zero or a positive integer.
Let us now write down the Schrödinger equation for an α-particle (z = 2) leaving a
recoiling nucleus (Z,A). For r > R
2 2
zZe 2

  (r ) 
 (r )  Q (r )
2M
40r
(21)
Here M is the reduced mass of the system. We do the usual separation of
variables: let
 (r )  R(r )Yl m (cos  ,  )
(22)
The spherical harmonic Y defines the orbital angular momentum l and its
z-component m for the outgoing α-particle.
Putting R (r) = U(r)/r and substituting into equation (21) the radial function U(r)
should satisfy the following equation:
 2 d 2U (r )  zZe 2 l (l  1) 2 



U (r )  QU (r )
2
2 
2M dr
2Mr 
 40r
zZe 2
40 r
l (l  1) 2
2Mr 2
(23)
The Coulomb barrier
The angular momentum barrier
It is clear that the total barrier is harder to penetrate and the transition rate will be
lower (and the mean life longer) than with the Coulomb barrier alone.
Blatt and Weisskopf (1952) have given some figures for the suppression
factors in α-decay transition rates due to the angular momentum barrier.
l
0
1
2
3
4
ωl/ω0
1.0
0.7
0.37
0.137
0.037
5
6
0.0071 0.0011
Values of the suppression factor due to the angular momentum barrier in
an α-decay for which Z = 86, Tα = 4.88 MeV, R = 9.87 fm
Note that we have assumed that the particle is emerging from the nucleus.
However, both the Coulomb and angular momentum barrier effects can apply also
to particles entering the nucleus.
And this is relevant to the rates of nuclear reactions where the first step is the
penetration into the nucleus by an incident particle.
~ The End ~
Rosetta stone,
offered the first step for modern
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