1000 Solved Problems in Modern Physics

188 3 Quantum Mechanics – II

Going back to (3)

1

Y

∂^2 Y

∂y^2

+

1

Z

∂^2 Z

∂z^2

=−α^2

1

Z

∂^2 Z

∂z^2

=−

1

Y

∂^2 Y

∂y^2

−α^2 (6)

Each side must be equal to a constant, say−γ^2 for the same argument as

before.

−

1

Y

∂^2 Y

∂y^2

−α^2 =−γ^2

Or

−

1

Y

∂^2 Y

∂y^2

+

(

α^2 −γ^2

)

= 0

Or

∂^2 Y

∂y^2

+μ^2 Y= 0

whereμ^2 =α^2 −γ^2 (7)

Y=Dsinμy

Y= 0 at y=b

This givesμ=

nyπ

b

(8)

Going back to (6)

(

1

Z

)

d^2 Z

dz^2

=−γ^2

This givesZ=Fsinγz

whereγ=

nzπ

c

(9)

∴ψ∼sin

(nxπx

a

)

sin

(nyπy

b

)

sin

(nzπz

c

)

(b) Combining (4), (5), (7), (8) and (9)

μ^2 =α^2 −γ^2 =(2mE/^2 )−β^2 −γ^2

Or

2 mE

^2

=μ^2 +β^2 +γ^2 =

(nyπ

b

) 2

+

(nxπ

a

) 2

+

(nzπ

c

) 2

Or

E=

(

h^2

8 m

)(

n^2 x

a^2

+

n^2 y

b^2

+

n^2 z

c^2

)

(10)

3.40 Fora=b=c
E=(h/ 8 ma^2 )

(

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

)

(Equation 10 of Prob 3.39)