半导体物理与器件第四版课后习题答案3

Chapter 3

3.1

If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and the

material would begin to behave more like an insulator.

_______________________________________ 3.2

Schrodinger's wave equation is: Assume the solution is of the form: Region I: ()0=x V . Substituting the

assumed solution into the wave equation, we obtain:

which becomes

This equation may be written as

Setting ()()x u x u 1= for region I, the equation becomes: where

In Region II, ()O V x V =. Assume the same form of the solution:

Substituting into Schrodinger's wave equation, we find:

This equation can be written as:

Setting ()()x u x u 2= for region II, this equation becomes where again

_______________________________________ 3.3

We have

Assume the solution is of the form: The first derivative is

and the second derivative becomes Substituting these equations into the differential equation, we find Combining terms, we obtain We find that

For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above.

_______________________________________ 3.4

We have the solutions for a x <<0 and for 0<<-x b .

The first boundary condition is which yields

The second boundary condition is which yields

The third boundary condition is which yields

and can be written as

The fourth boundary condition is which yields

and can be written as

_______________________________________ 3.5

(b) (i) First point: πα=a

Second point: By trial and error, (ii) First point: πα2=a

Second point: By trial and error,

_______________________________________ 3.6

(b) (i) First point: πα=a

Second point: By trial and error, (ii) First point: πα2=a

Second point: By trial and error,

_______________________________________ 3.7

Let y ka =, x a =α Then

Consider dy

d

of this function.

We find Then

For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general, And So

This implies that

dk dE dk d =

=0α for a

n k π= _______________________________________ 3.8

19104114.3-⨯=J From Problem 3.5 18100198.1-⨯=J 19107868.6-⨯=J

or 24.4106.1107868.619

19

=⨯⨯=∆--E eV

18103646.1-⨯=J From Problem 3.5, 18103364.2-⨯=J

1910718.9-⨯=J

or 07.6106.110718.919

19

=⨯⨯=∆--E eV

_______________________________________ 3.9

(a) At π=ka , πα=a 1 19104114.3-⨯=J