1000 Solved Problems in Modern Physics

(Romina) #1

3.3 Solutions 183


G=

2



(2m)

1
2

∫b

a

(

zZe^2
r

−E

) 1 / 2

dr

wherez= 2
Now at distancebwhere the alpha energy with kinetic energyE, potential
energy=kinetic energy

E=

1

2

mv^2 =zZe^2 /b

G=

(

2



)

(

2 mzZe^2

) 1 / 2 ∫b
a

(

1

r


1

b

) 1 / 2

dr

The integral is easily evaluated by the change of variabler=bcos^2 θ

I=


b

{

cos−^1

(a
b

)



a
b


a^2
b^2

}

Finally

G=

2



(

2 mz Z e^2 b

) 1 / 2

[

cos−^1

(

R

b

)



R

b


R^2

b^2

]

wherea=R, the nuclear radius.
Ifvinis the velocity of the alpha particle inside the nucleus andR=ais
the nuclear radius then the decay constantλ= 1 /τ∼(vin/R).e−G
3.35 (a) In Problem 3.19 the condition that a bound state be formed was obtained as

cotkR=−

γ
k

=−

[

W

V 0 −W

] 1 / 2

whereV 0 is the potential depth andais the width. Here the condition
would read

cotka=−

[

W

V 0 −W

] 1 / 2

wherek^2 = 2 m(V 0 −W)a^2 /^2
If we now makeW=0, the condition that only one bound is formed is

ka=

π
2

orV 0 =

h^2
32 ma^2

(b) The next solution is
ka=

3 π
2
Here W 1 =0 for the first excited state
With the second solution we get

V 1 =

9 h^2
32 ma^2
Note that in Problem 3.23 the reduced massμ=M/2 while hereμ=m.
The graphs are shown in Fig. 3.13.
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