Solution manual Fundamentals of Semiconductor Devices, 2nd Edition, by Anderson (COMPLETE)

Chapter 1 problems

1.1 Show that Equation (1.6) follows from Equation (1.3).
q2 dr
Equation 1.3: dEP = dEP ( r ) = -Fdr =
4pe 0 r 2
We know that EP ( r = ¥ ) = Evac (Equation 1.4)
Evac ¥ q2 dr
Substituting in: òEP
dE P = ò
r 4pe 0 r 2
(Equation 1.5)


Carrying out the integrals:
q2 dr ¥ dr
4pe 0 òr r 2
Evac - EP =

q2 é 1 ù
¥

= ê- ú
4pe 0 ë r r û
q2 é 1ù q2
=- 0 - = +
4pe 0 êë r úû 4pe 0 r
q2
EP = Evac -
4pe 0 r


1.2 Fill in the steps to derive Equations (1.12) and (1.13) from Equations (1.8)
and (1.11.) (This is corrected)
mv2 q2
Equation 1.8: - =0
r 4pe 0 r 2
Equation 1.11:
Taking the first and multiplying by rn2 :
mvn 2 2 q2
rn − rn 2 = 0
rn 4 0 rn 2
q2
mvn 2 r =
4 0
q2
( mvn rn ) vn =
4 0
q2
( n฀ ) vn =
4 0
q2
vn =
4 0 ( n฀ )
That is Equation 1.13. Now substitute in this expression for vn in Equation 1.8:

, mvn2 q2
− =0
rn 4 0 rn2
2
m  q2  q2
  =
rn  4 0 n฀  4 0 rn2
m  q2 
  =1
rn  4 0 n฀ 
 4 0 n 2 ฀ 2 
 2  = rn
 mq 
which is Equation 1.12.

1.3 Using the Bohr model, find the first three energy levels for a He+ ion, which
consists of two protons in the nucleus with a single electron orbiting it. What
are the radii of the first three orbits?

We repeat the analysis that we used for the Hydrogen atom, except that the charge
on the nucleus is nowQ1=+2q.

Q1Q2 -2q2
F= =
4pe 0 r 2 4pe 0 r 2

2q2
EP (r ) = Evac -
4pe 0 r

mv2 2q2
- =0
r 4pe 0 r 2

, Thus E1 = Evac - 4(13.6eV) = Evac - 54.4eV

and E2 = Evac - 4(13.6eV) / 22 = Evac -13.6eV

and E3 = Evac - 4(13.6eV) / 32 = Evac - 6.04eV



1.4. For each of the potential energy distributions in Figure 1P.1, sketch the
force on an electron in this region of space, paying attention to magnitude and
sign.
We know that force is minus the gradient of the potential F = -dEP / dr , so we need
to sketch the negative slope.

, 1.5) Consider the electron in the energy diagram in Figure 1P.2. It is initially at
point A , and moving to the left. At point C, it collides with an atom and loses
some energy.




a) What is its initial potential energy?
The potential energy in this region is E0.

b) What is its initial kinetic energy?
The total energy is 0.1 eV above the potential energy at that point, so the difference,
the kinetic energy, is 0.1 eV.

c) What is its initial acceleration?
The slope of EP =0 in this region, so there is no force and thus no acceleration.

d) At point B, what is its potential energy? Kinetic energy?
At B the potential energy EP has decreased to about E0-0.1 eV. The total energy has
remained the same by conservation of energy, or Etotal=E0+0.1eV. The difference
between the total energy and the potential energy is the kinetic energy, or
EK=E0+0.1 –(E0-01)=0.2 eV.

e) What are its potential and kinetic energies just before the collision?
Just before the collision, the potential energy is about E0-0.35 eV, and the kinetic
energy is therefore EK=Etotal-EP=E0+01-(E0-0.35)=045 eV.

f) What are its potential and kinetic energies just after the collision?
After the collision, the electron has lost some energy. Its total energy is now E0-0.2
eV, the potential energy is still E0-0.35 eV , so the kinetic energy is +0.15 eV.

g) Is the electron still moving after the collision?
It still has kinetic energy so it has to be moving.

h) Where did the energy lost by the electron go?
It was given to the atom that it hit.