Alpha Decay II
[Sec. 7.1/7.2/8.2/8.3 Dunlap]
The one-body model of α-decay assumes that the
α-particle is preformed in the nucleus, and confined
to the nuclear interior by the Coulomb potential
barrier. In the classical picture, if the kinetic energy
of the -particle is less than the potential energy
represented by the barrier height, the α-particle
cannot leave the nucleus.
vα
Qα
r=R
  
    PT
R 
r=b
In the quantum-mechanical picture, however, there is
a finite probability that the -particle will tunnel
through the barrier and leave the nucleus.
The α-decay constant is then a product of the
frequency of collisions with the barrier, or ``knocking
frequency'‘ (vα/2R), and the barrier penetration
probability PT.
How high and wide the barrier?
2Ze 2
1
V (r ) 
 2Z .c.
(4 0 )r
r
The height of the barrier is:
Emax
2 Z . .c

R
The width of the barrier is
30MeV
w
Lets calculate these for
Emax
w  bR 
235
92
U
taking R0=1.2F, we have
2 x 92 x197 MeV .F

 36 MeV
137 x 7.4 F
2Z . . c
R
Q
R  1.2x(235)1/ 3  7.4F
2x92 x197 MeV .F
w
 7.4 F  49 F
137 x 4.68 MeV
The 1D square potential tunneling problem
This problem is quite algebraically difficult to solve exactly
(although it can be done) because one has to match the
sinusoidal wavefunctions and the gradient of the wavefunction at
barrier entry (A) and output (B).
The 1D square potential tunneling problem
The 3D tunneling problem
So taking the case when l=0, we have to solve the 1D SE for
the potential
The WKB approximation
How do we apply this result to the case of
the barrier with a height V, that is not
constant? As with the finite barrier really
exact solution is not possible, but a good
approximation exists – it is called the WKB
approximation- after physicists (Wentzel,
Kramers and Brillouin).
The WKB approximation
which in the limit of r
becoming small becomes:
G is known as the Gamow factor. It is
this Gamow factor that it is now our
task to calculate.
The WKB approximation
(Q in MeV and R in fm)
The WKB approximation
The formula for decay rate
So finally we have to put everything
together to get the decay rate of an
alpha unstable nucleus. The decay rate
can be considered as a compounding of
three probabilities.
(i) The probability Pα of the α-particle
forming (or being in existence) at a
specific time
(ii) The probability fα of the α-particle
hitting the “wall” of the nucleus each
second
(iii) The probability T of the α-particle
transiting to the “freedom distance” r=b
The formula for decay rate
where the Gamow factor G can be
either expressed by (8.17) or by (1)
depending on preference.
Explaining the Geiger-Nuttal Law
c 2(Q  U ) G
e
  P .
2
mc
R
1/ 2
 0e
 Z 
3.95 
 Q 
 2.97( ZR )1/ 2
Where we have ignored the small
pre-exponential dependence on Q.
We find on taking logarithms
1/ 2
 Z 
ln   ln 0  2.97( ZR)  3.95  
 Q 
1/ 2
The range energy relationship for alphas is
Nuttal law gives
R  T3/ 2
so that the Geiger-
ln   ln Q . The barrier penetration theory gives ln   1/ Q
In the range 4<Q<7MeV which covers most alpha emission – a quantity linear in
is close to be linear in 1/ Q so that the Geiger Nuttal law is vindicated.
ln Q