Selected old midterm exams
1. A 0.5-cm solid sample of rectangular cross section area 1mm x 2mm has a
resistance R = 600 Ω. In a Hall effect measurement, a Hall field of 1.2 V/m
is detected in the sample when a current of 3 mA flows through and a
magnetic field of 100 G is applied in a direction normal to the current flow.
a)
Calculate the value of the charge carrier density in this sample
b)
Find the value of the mobility for the charge carrier and specify its
units.
Soln
(a)
Ne 

1
BJ
BI


RH q EH q qEH A
(100 104 )(3 103 )
(1.6 1019 )(1.2)(1 2 10 4 )
N e  7.811011 cm -3
(b)
J   E  ( N e q e ) E 
( N e qe )V ( N e q e )( IR)

L
L
I ( N e qe )( IR)

A
L
L
0.5 102
e 

AN e qR (1 2 104 )(7.811017 )(1.6 1019 )(600)
e  0.333 m 2 /V.s  3.333 103 cm 2 /V.s
2. A Hall effect experiment is shown below with d = 0.2 cm, w = 0.1 cm, I = 5
mA, BZ = 1 W/m2, V = 0.245 V, and |RH| = 1.18 cm3/C. A voltage meter is
connected to read VH with positive charges on the top as shown in the
figure. Find (a) type of this material (n- or p-type). Please explain. (b)
majority carrier concentration (c) Hall voltage VH
Soln
(a) The force due to Hall effect will be in – y direction. Therefore, this is an
n-type material.
(b)
RH  
n
(c)
1
qn
1
1.6  10
19
1.18
 5.297  1018 cm-3
VH  EHW
 R . I .B 
 H
W
 A 
 R . I .B 
 H
W
 W .d 
1.18  5  10 3  10 2

0.1
VH  5.9 V
3. (a) Find the resistivities of intrinsic Si and intrinsic GaAs at the room
temperature.
(b) Find the room-temperature resistivities of an n-type Si doped with ND
= 1016 cm-3.
Soln
(a) For intrinsic Si, n = 1450, p = 505, and n = p = ni = 9.65109 cm-3
1
1
We have  

 3.31105 -cm
qnn  qp p qni (n   p )
(b) From the figure, μn = 1300 cm2/V-s

1
1

 0.48 -cm.
19
qnn 1.6  10  1016  1300
4. Determine the conductivities of intrinsic Si and intrinsic GaAs at room
temperature.
Note: For intrinsic Si, n = 1450, p = 505, and n = p = ni = 9.65109 cm-3 .
Similarly for GaAs, n = 9200, p = 320, and n = p = ni = 2.25106 cm-3
Soln
 Si 
1
1

 3.31105
qnn  qp  p qni (  n   p )
 Si 
1

 3.021 106 cm -1. -1
GaAs 
1
1

 2.92  108
qnn  qp  p qni ( n   p )
 GaAs 
1

 3.425  109 cm -1. -1
5. An electron and a photon have the same value for energy, but the
wavelength of the photon is a hundred times the wavelength of the
electron.
a) What is the value of their common energy?
b) Calculate the values of those two wavelengths.
Soln
(a)
h
h

p mv
1
1h
Ee  mv 2 
v
2
2
hc
E ph  h ph 

 ph
Ee  E ph
1h
hc
v
2
 ph
1v
c

2  100
2c
v
 6 106 m/s
100
1
Ee   9.11 1031  (6  106 ) 2  1.639  1017 J = 1.02  102 eV
2
(b)
h
6.62 1034

 1.2111010 m
31
6
mv 9.1110  6 10
 ph  100e  121.11010 m
e 
6. Electromagnetic radiation of wavelength 20 nm is incident on atomic
hydrogen. Assuming that an electron in its ground state is ionized what is
the maximum velocity at which it may be emitted?
Soln
13.6
eV
n2
E  E1  13.6 eV
En  
radiation energy 
hc


6.63 1034  3 108
 62.06 eV
20  109  1.6  1019
1 2
mv
2
48.46  2 1.6 10 19
v2 
9.1 1031
v  4.128 106 m/s
62.06  13.6 
7. An electron in a ground state E1 of a potential well is excited to a higher
energy states E2 by a photon with wavelength of 560 nm. Another electron
in E2 is excited by a photon with wavelength of 920 nm to E3. Find the
energy E3 of this potential well in eV if E1 = 0 eV.
Soln
E2  E1  h 12
E2 
hc
12
0
6.63 1034  3 108
560 109 1.6 1019
E2  2.217 eV

E3  E2  h 23
E3 
hc
23
 2.217
6.63 1034  3 108
 2.217
920 109 1.6  1019
E3  3.566 eV

8. Find the intercepts of these following Miller indices
(1 0 0)
(1 1 0)
(6 4 3)
Soln
Do the inverse procedure of how to get Miller Indices.
(a) (1 0 0)
Reciprocals: (1  )
Interceptions: (1  )
(b) (1 1 0)
Reciprocals: (1 1 )
Interceptions: (1 1 )
(c) (6 4 3)
Reciprocals: (1/6 1/4 1/3)
Multiply by LCM that is 12: (2 3 4)
Interceptions: (2 3 4)
9. (a) If a plane has intercepts at 2a, 3a, and 4a along the three Cartesian
coordinates, where a is the lattice constant, find the Miller indices of the
planes.
(b) By consider their melting points, is Si or GaAs easier to break the
bond?
Soln
(a) Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. The
smallest three integers having the same ratio are 6, 4, and 3. The plane is
referred to as (643) plane.
(b) Melting point of Si and GaAs is 1,412C and 1,240C, respectively.
Thus, it is clearly seen that it is easier to break Ga-As bond than Si-Si
bond.
10. Calculate the number of atoms per unit cell and the closest possible
distance between two atoms in the following cubic structures by assuming
lattice constant a = 4 Å
a) Simple Cubic (SC)
b) Body-Centered Cubic (BCC)
c) Face-Centered Cubic (FCC)
d) Diamond Structure
e) Zinc Blende Structure
Soln
(a) # of atoms/unit cell = 8 x (1/8) = 1
2r = a = 4 Å
(b) # of atoms/unit cell = [8 x (1/8)] + 1 = 2
2r 
(c) # of atoms/unit cell = [8 x (1/8)] + [6 x (1/2)] = 4
2r 
a 2
 2.828
2
d) # of atoms/unit cell = 8
2r 
(e) same as (d)
a 3
 1.732 Å
4
Å
a 3
 3.464 Å
2
11. (a) Calculate the number of Si atoms per cubic centimeter and the density
of Si at room temperature.
(b) Repeat (a) if Si is replaced by GaAs
Soln
(a) There are eight atoms per unit cell.
Therefore, 8 / a 3  8 /  5.43 108   5 1022 cm -3
3

nM
5  1022  28.09

 2.333 g/cm3
N Aa3
6.02 1023
(b) There are eight atoms per unit cell as well for GaAs.
Therefore, 8 / a 3  8 /  5.65 108   4.436 1022 cm -3
3
nM
4.436  1022  144.63


 10.66 g/cm3
3
23
N Aa
6.02  10
12. Calculate the atoms surface density (# of atoms/cm2) for Ga atoms on a Ga
terminated (001) surface in GaAs. The lattice constant for GaAs is 5.65 Å.
Soln
1 
There are   4  +1 = 2 atoms in (001) surface.
4 
Therefore, 2 / a 2  2 /  5.65 108   6.265 1014 cm -2
2
13. A Unit cell of NaCl is shown below. Find a) number of atoms per unit cell
of Na b) number of atoms per unit cell of Cl c) Atom surface density of Cl
on (010) surface if a = 5 Å
Soln
(a)
8 ions at the corners and 6 on the surface (Na)
 1  1
# of Na atoms/unit cell =  8     6    4
 8  2
(b)
12 ions at the center and 1 in the center of a cell
1

# of Na atoms/unit cell = 12    1  4
4

(c )
# of Cl at (0 1 0) = 4 
1
2
2
Atom surface density =
2
 5 10 
-10 2
 8 1018 m -2
14. (a) Calculate the density of GaAs where atomic weights of Ga and As are
69.72 and 74.92 g/mol, respectively.
(b) If GaAs is doped with Si and Si has replaced Ga in a crystal, will it be
‘p-’ or ‘n-’ type semiconductor?
Soln
(a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga
and As are 69.72 and 74.92 g/mole, respectively. There are four gallium
atoms and four arsenic atoms per unit cell, therefore
4/a3 = 4/ (5.65 × 10-8)3 = 2.22 × 1022 Ga or As atoms/cm2,
Density = (no. of atoms/cm3 × atomic weight) / Avogadro constant
= 2.22 × 1022(69.72 + 74.92) / 6.02 × 1023 = 5.33 g / cm3.
(b) If GaAs is doped with Si and Si atoms displace Ga atoms, donors are
formed, because Si has four valence electrons while Ga has only three.
The resulting semiconductor is n-type.
15. In a one dimensional metal model, the electron density per unit length is
6.52x107cm-1. Calculate the Fermi-energy level (EF) in eV at zero absolute
temperature.
Soln
n   g ( E ) F ( E )dE
For T = 0K
n
EF
2m


0

n
2m

E 1/ 2 dE
EF
E
1/ 2 0
2 2mEF

EF (T  0 K ) 
n 2 2
8m
2
 6.52 10

9
  1.05 1034 
8  9.111031
 6.347 1019 J = 3.967 eV
2
16. What is the probability that a state at 1 eV above the Fermi level will be
occupied when temperatures is -273°C, -196°C, 27°C, 727°C?
Soln
F (E) 
1
1 e
( E  EF ) / kT
17. A Si sample is doped with 1017 As atoms/cm3. What is the equilibrium
hole concentration p at 300K.
Soln
n  N D  1017 cm-3
ni2  9.65  10
p

ND
1017

9 2
 931.23 cm-3
18. At room temperature, a Si sample is doped with 1016 Phosphorus
atoms/cm3. Find:
(a) the equilibrium majority carrier concentration (ionized atoms)
(b) the equilibrium minority carrier concentration
(c) EC – ED and EC - EF in eV
Soln
(a) At room temperature, all donor atoms are ionized.
(b)
(c)
EC – ED = 0.045 eV
EC – EF = 0.2061 eV
19. At room temperature, a Si sample is doped with 1016 Boron atoms/cm3.
Find:
(a) the equilibrium majority carrier concentration (ionized atoms)
(b) the equilibrium minority carrier concentration
(c) EA – Ev and EF - EV in eV
Soln
(a) At room temperature, all acceptor atoms are ionized.
p  N A  1016 cm-3
(b)
(c)
ni2  9.65  10
n

NA
1016

9 2
 9312 cm-3
E A  EV  0.045 eV
 NV 
 2.66  1019 
EF  EV  kT ln 
  0.0259ln 
  0.204 eV
16
 10

 NA 