1000 Solved Problems in Modern Physics

3.3 Solutions 179

α^2 a^2 =

2 mEa^2

^2

=

n^2 π^2

4

E=

n^2 π^2 ^2

8 ma^2

(neven)

ThusE=

n^2 π^2 ^2

8 ma^2

,n= 1 , 2 , 3 ...

3.28 From problem 3.27,

αcotαa=−β= 0

Thefirstsolutionisαa=π/2, for the ground state.

The second solution,αa= 3 π/2, will correspond to the first excited state

(withl=0). This will give

α^2 a^2 = 9 π^2 / 4

Let the excited states be barely bound so thatW=0. Then,

α^2 = 2 mE/^2 = 9 π^2 / 4 a^2

E=V 1 = 9 V 0

a value which is not possible. Thus, the physical reason why bound excited

states are not possible is that deuteron is a loose structure as the binding energy

(2.225 MeV) is small. The same conclusion is reached for higher excited states

includingl= 1 , 2 ...

3.29 The inside wave functionu 1 =Asinkris maximum atr≈R. Therefore

kR=

π

2

ork^2 =

M(V 0 −W)R^2

^2

=

π^2

4

orV 0 =

(

π^2 ^2 c^2

4 Mc^2 R^2

)

+W

Substitutingc= 197 .3MeV−fm,Mc^2 =940 MeV,R= 1 .5 fm andW=

2 .2 Mev, we findV 0 ≈47 MeV

3.30 Schrodinger’s equation in one dimension

(a)d

(^2) ψ
dx^2 +
( 2 m
^2

)

(E−V)ψ= 0

Region 1: (x<0)V=0;d

(^2) ψ
dx^2 +k
2
1 ψ=^0
wherek^21 =^2 mE 2 Solution:ψ 1 =exp(ik 1 x)+Aexp(−ik 1 x)
Region 2: (0<x<a)V=Vb;d
(^2) ψ
dx^2 −k
2
2 ψ=^0
wherek^22 =
( 2 m
^2

)

(Vb−E)

Solution:ψ 2 =Bexp(k 2 x)+Cexp(−k 2 x)