bei48482_FM

Theory of Alpha Decay 469

and then express it as the integral

ln T 2 

L

0

k 2 (r) dr 2 

R

R 0

k 2 (r) dr (12.34)

where R 0 is the radius of the nucleus and Ris the distance from its center at which

UE. The kinetic energy Eis greater than the potential energy Ufor rR, so if it

can get past R, the alpha particle will have permanently escaped from the nucleus.

The electric potential energy of an alpha particle at the distance rfrom the center

of a nucleus of charge Zeis given by

U(r)

Here Zeis the nuclear charge minusthe alpha-particle charge of 2e; thus Zis the atomic

number of the daughter nucleus.

We therefore have

k 2 

1  2

 E

1  2

Since UEwhen rR,

E (12.35)

and we can write k 2 in the form

k 2 

1  2

 ^1 

1  2

(12.36)

Hence

ln T 2 

R

R 0 k^2 (r) dr

 (^2) 
1  2

R
R 0  ^1 
1  2
dr
 (^2) 
1  2
R cos^1 
1  2
  
1  2
^1  
1  2
(12.37)
Because the potential barrier is relatively wide, RR 0 , and
cos^1 
1  2
 
1  2
^1  
1  2
 1
with the result that
ln T (^2) 
1  2
R  (^2) 
1  2
R 0

R

2
2 mE

^2
R 0

R
R 0

R

2
R 0

R
R 0

R
R 0

R
R 0

R
2 mE

^2
R

r
2 mE

^2
R

r
2 mE

^2
2 Ze^2

4

0 R
2 Ze^2

4

0 r
2 m

^2
2 m(UE)


2 Ze^2

4

0 r
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