1000 Solved Problems in Modern Physics

3.3 Solutions 195

When the boundary conditions are imposed,

β=

n 2 πy

a

ψy=Gsin

(n 2 πz

a

)

Thusψ(x,y)=ψxψy=Ksin

(n 1 πx

a

)

sin

(n 2 πy

a

)

(K=constant)

andα^2 +β^2 = 2 mE/^2 =

(n 1 π

a

) 2

+

(n 2 π

a

) 2

or

E=

(

^2 π^2

2 ma^2

)(

n^21 +n^22

)

3.46 By Problem 3.39,E=

(

h^2

8 ma^2

)(

n^2 x+n^2 y,n^2 z

)

Therefore the numberNof states whose energy is equal to or less thanEis

given by the condition

n^2 x+n^2 y+n^2 z≤

8 ma^2 E

h^2

The required number,N=

(

n^2 x+n^2 y+n^2 z

) 1 / 2

, is numerically equal to the

volume in the first quadrant of a sphere of radius

(

8 ma^2 hE 2

) 1 / 2

. Therefore

N=

(

1

8

)

·

(

4 π

3

)(

8 ma^2 E

^2

) 3 / 2

=

2 π

3

(

ma^2 E

2 ^2 π^2

) 3 / 2

3.47 Schrodinger’s radial equation for spherical symmetry andV=0is

d^2 ψ(r)

dr^2

+

2

r

dψ(r)

dr

+

2 mE

^2

ψ(r)= 0

Take the origin at the centre of the sphere. With the change of variable,

ψ=

u(r)

r

(1)

The above equation simplifies to

d^2 u

dr^2

+

2 mEu

^2

= 0

Thesolutionis

u(r)=Asinkr+Bcoskr

wherek^2 =

2 mE

^2

(2)

Boundary condition is:u(0)=0, becauseψ(r) must be finite atr=0. This

givesB= 0

Therefore,

u(r)=Asinkr (3)

Furtherψ(R)=u(Rr)= 0

SinkR= 0