1000 Solved Problems in Modern Physics

(Tina Meador) #1

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
Free download pdf