1000 Solved Problems in Modern Physics

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.