Banha University

Banha University First Term 2012/2013
Faculty of Engineering Regular Exam
Department of Basic Sciences Students: 1 ST Written Examination
Year El. Eng.
Solutions to Questions for the First-Term Examination (January 2013)
Subject: Modern Physics S1133 Time: 3 Hours
Attempt all questions. No. of questions: 5 No. of pages:3
Solution to Question1 (18 points)
a) The wave-function that satisfies Schrodinger equation for an electron, in onedimensional
infinite potential well of length a = 1nm is
Ψ(x) =
2 3x
sin
a a
(i) What is the energy of the electron in this state? (ii) What is the de Broglie
wavelength of the electron? (iii) What is the probability of finding the electron
between x = a/2 and x = 2a/3?
Solution
(i) From the wave-function k =
In one-dimensional infinite potential well V = 0 for 0 < x < a.
The energy of the electron E =
(ii) Since E =
Then the momentum p =
The de Broglie wavelength λ =
(iii) The probability density P(x) = ψ 2 (x) =
The probability p(a/2 < x

Solution
Bohr model: the hydrogen atom consists of a positively charged nucleus
containing a proton and an orbiting electron.
Bohr postulates:
1. The electron moves in stable circular orbits around the proton under the influence
of the electric force of attraction
2. The allowed orbits are those for which the electron’s orbital angular momentum
about the nucleus is quantized and equal to an integral multiple of ħ: mvr = nħ
where m is the electron mass, v is the electron’s speed in its orbit.
3. The atom emits radiation when the electron makes a transition from a
more energetic initial stationary state to a lower-energy stationary state. The
frequency of the emitted radiation is hf = Ei – Ef
where Ei is the energy of the initial state, Ef is the energy of the final state,
and Ei > Ef.
The radius of the orbits:
The electric force exerted on the electron must equal the product of its mass and its
centripetal acceleration:
Using Bohr’s postulate: mvr = nħ
The energy of the electron
The kinetic energy of the electron
Substituting for the radius of the orbits, then then the total energy

where
The emitted wavelength of light:
Where RH =
The four quantum numbers (n, l, ml, ms) are:
The first quantum number is the principal quantum number n. It is associated with
the energies of the allowed states for the hydrogen atom
The values of n are integers that can range from 1 to ∞.
The second quantum number is the orbital quantum number l. It is associated with
the orbital angular momentum of the electron.
The values of l are integers that can range from 0 to (n – 1).
The third quantum number is the orbital magnetic quantum number ml. It specifies
the allowed values of the z component of the orbital angular momentum L.
LZ = ml
The values of ml are integers that can range from - l, …, -2, -1, 0, 1, 1,…, l.

L lies anywhere on the surface of a cone that makes an angle θ with the z axis such
that
The fourth quantum number is the spin magnetic quantum number ms. There are
only two directions to exist for the electron spin. Either the spin is up or down
relative to the z-axis. The values of the spin magnetic quantum number are
ms = - ½, + ½ .
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Solution to Question 2 (18 points)
a) The Fermi energy for copper is 7.0 eV. (i) Find the probability of an energy level
at 7.15 eV is being occupied by an electron at 300K if the electrons in copper follow
the Femi-Dirac distribution function. (ii) Find the Fermi level at 0K if the density of
electrons 8.75x10 22 cm -3 .
Solution
(i) The probability f(E) = ( E EF ) / kT
1
e
For E = 7.15 eV, EF = 7 eV, kT = 0.026 eV
f(E = 7.15 eV) = ( 7.
15 7)
/ 0.
026
1
e
(ii) The electron density n = g ( E)
f ( E)
dE
0
The density of states g(E) =
1
8
2m
3
h
3/
2
1
E
1/
2
The probability at T = 0K: f(E) = 1 for E < EF
= 0 for E > EF
8
Then n =
3/
2 EF
2m
1/
2 2 8
E dE
3
h
3
3/
2
2m
3/
2
E 3 F
h
0
2 / 3
= 3.11x10 -3 = 0.311 %
2
h 3n
The Fermi level EF(0) =
2m
8
34
2
22 6
2
( 6.
63x10
) 3(
8.
75x10
x10
)
31
2(
9.
1 10 )
8
x
1.
154 x10 -18 = 7.213 eV
J
2 / 3

) Derive expressions for the position of Fermi energy level in each of the following
(i) intrinsic semiconductor,
(ii) n-type semiconductor, and
(iii) p-type semiconductors.
Sketch this position in the energy band diagram for each of semiconductor type.
Solution
(i) The position of Fermi level in intrinsic semiconductor:
Since the electron and hole concentrations are equal,
then no = po
(ii) The position of Fermi level in n-type semiconductor:
For n-type no = Nd, then

Also
EF – EFi = kT ln(Nd/ni)
(iii) The position of Fermi level in p-type semiconductor:
For p-type po = Na, then
Also
EFi – EF = kT ln(Na/ni)
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Solution to Question 3 (18 points)
a) Three samples of silicon at T = 300 K are the following:
Sample 1 is n-type doped with arsenic atoms to Nd = 5x10 16 cm -3 .
Sample 2 is p-type doped with boron to Na = 2x10 16 cm -3 .

Sample 3 is compensated silicon doped with both donors and acceptors
described in the n-type and p-type samples.
Find (i) the electron and hole concentrations in each sample, (ii) the conductivity of
each sample (iii) The current density passing in each sample if an electric field of 25
V/cm is applied.
Solutions
(i) sample 1: no = Nd = 5x10 16 cm -3
Po = ni 2 /Nd = (1.5x10 10 ) 2 /5x10 16 = 4.5x10 3 cm -3
sample 2: po = Na = 2x10 16 cm -3
no = ni 2 /Na = (1.5x10 10 ) 2 /2x10 16 = 1.125x10 4 cm -3
sample 3: no = (Nd - Na)/2 + {[(Nd - Na)/2] 2 + ni 2 } 1/2
no = (Nd - Na)/2
= (5x10 16 – 2x10 16 )/2 = 1.5 x10 16 cm -3
Po = ni 2 /no = (1.5x10 10 ) 2 /1.5x10 16 = 1.5x10 4 cm -3
(ii) sample 1: σn = e no μn = (1.6x10 -19 )(5x10 16 )(1350) = 10.8 S/cm
sample 2: σp = e po μp = (1.6x10 -19 )(2x10 16 )(480) = 1.536 S/cm
sample 3: σn = e no μn
= (1.6x10 -19 )(1.5x1016)(1350) = 3.24 S/cm
(iii) sample 1: Jn = σn ε = 10.8 x 25 = 270 A/cm 2
sample 2: Jp = σp ε = 1.536 x 25 = 38.4 A/cm 2
sample 3: Jn = σn ε = 3.24 x 25 = 81 A/cm 2
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b) (i) Describe how the space charge region is formed in pn junction diode.
(ii) Derive an expression for the electric field and the electric potential in the pn-
junction diode.
(iii) Sketch the energy band diagram of zero-biased, reverse-biased and forward-
Biased pn-junction.
Solutions
(i) Formation of space charge region in pn-junction diode:
A single-crystal semiconductor material in which one side is doped with acceptor
impurity atoms to form the p region and the the other side is doped with donor
atoms to form the n region. The interface separating the n and p regions is referred
to as the junction.
Initially, at the junction, there is a very large density gradient in both the electron
and hole concentrations. Majority carrier electrons in the n region will begin
diffusing into the p region leaving behind positively charged donor ions and majority
carrier holes in the p region will begin diffusing into the n region leaving behind

negatively charged acceptor ions. If we assume there are no external connections
to the semiconductor, then this diffusion process cannot continue indefinitely. The
density gradient produces a "diffusion force" that acts on the majority carriers.
These diffusion forces, acting on the electrons and holes at the edges of the space
charge region. The net positive and negative ion charges in the n and p regions
induce an electric field in the region near the junction, in the direction from the
positive to the negative charge, or from the n to the p region. The electric field in
the junction region produces another force on the electrons and holes which is in
the opposite direction to the diffusion force for each type of particle. In thermal
equilibrium, the diffusion force and the E-field force exactly balance each other.
The net positively and negatively charged regions are referred to as the space
charge region. Essentially all electrons and holes are swept out of the space charge
region by the electric field. Since the space charge region is depleted of any mobile
charge, this region is also referred to as the depletion region.
The Fermi energy level is constant throughout the entire system. The conduction
and valance band energies must bend as we go through the space charge region.
(ii) The electric field:
The charge density ρ(x) = - eNa - xp ≤ x ≤ 0
= e Nd 0 ≤ x ≤ xn

The electric field is assumed to be zero in the neutral p region for x < - xp since the
currents are zero in thermal equilibrium.
The electric field in p-region - xp ≤ x ≤ 0;
The electric field in the n-region 0 ≤ x ≤ xn,
The potential in p-region - xp ≤ x ≤ 0;
The potential in the n-region 0 ≤ x ≤ xn,

(iii) Energy band diagram
Zero-biased
Reverse-biased
Forward-biased
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Solution to Question 4 (18 points)
a) Consider a silicon pn-junction diode with a cross-sectional area of 10 -4 cm 2 has
the following properties at 300 K:
p-region n-region
Na = 5x10 15 cm -3
Nd = 10 17 cm -3
τn = 10 -6 s
Dn = 25 cm 2 /s
τp = 10 -7 s
Dp = 10 cm 2 /s
Calculate (i) the built-in voltage barrier Vbi, (ii) the space charge width, the
maximum electric field and the junction capacitance of the junction under reverse
voltage of 5 V, (iii) The ideal reverse saturation current Is, and the diode current at
a forward-bias voltage of 0.65 V.
Solution
(i) The built-in voltage Vbi = Vt ln (NaNd/ni 2 )
= (0.026) ln[(5x10 15 x10 17 )/(1.5x10 10 ) 2 ]
= 0.739 V
(ii) The space charge width W =
The maximum electric field
εmax =
2
s ( V bi V
R ) N a N
e
N a N d
=
14
15 17
2(
11.
7)(
8.
85x10
)( 0.
739 5))
5x10
10
19
15 17
1.
6x10
5x10
x10
= 1.25x10 -3 cm
εmam = 2(Vbi +VR)/W=2(0.739 +5)/( 1.25x10 -3 )=9.182x10 3 V/cm
The junction capacitance C = εsA/W = (11.7)(8.85x10 -14 )(10 -4 )/(1.25x10 -3 )
= 8.2836x10 -14 F
(iii) The ideal reverse saturation current
eD p
Lp
eD
p no n po
Is = A
L
n
n
Pno = ni 2 / Nd = (1.5x10 10 ) 2 /10 17 = 2.25x10 3 cm -3
Lp = √τpDp = √10 -7 x10 = 10 -3 cm
npo = ni 2 / Na = (1.5x10 10 ) 2 /5x10 17 = 450 cm -3
Ln = √τnDn = √10 -6 x25 = 5x10 -3 cm
d

10x2.
25x10
10
25x450
5x10
3
19
Is = 1. 6x10
3
3
4
16
2
x10
3.
96x10
A/
cm
The diode current I = Is(e V/Vt – 1)
= 3.96x10 -16 (e 0.65/0.026 – 1) = 2.85x10 -5 A
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b) Describe the operation, the charge flow and the current components in npn
bipolar junction transistor in the forward active mode. What are the operating
modes of the transistor?
Solution
The bipolar transistor consists of three separately doped regions and two pn
junctions. There are two types of transistor structures an npn bipolar transistor and
a pnp bipolar transistor. The three terminal connections are called the emitter,
base, and collector. The width of the base region is small compared to the minority
carrier diffusion length.
The operation of an npn transistor in the forward active mode and the
charge flow
The base-emitter (B-E) pn junction is forward-biased, and the base-collector
(B-C) pn junction is reverse-biased in the forward-active operating mode. Since the
B-E junction is forward-biased so electrons from the emitter are injected across the
B-E junction into the base. Also holes from the emitter are injected across the B-E
junction into the emitter.
The injected electrons create an excess concentration of minority carriers in the
base. The B-C junction is reverse biased, so the minority carrier electron
concentration at the edge of the B-C junction is ideally zero. The large gradient in
the electron concentration means that electrons injected from the emitter will
diffuse across the base region into the B-C space charge region where the electric
field will sweep the electrons into the collector. Some of the injected electrons
recombine with the majority holes in the base. We want as many electrons as

possible to reach the collector without recombining with any majority carrier holes
in the base. For this reason, the width of the base needs to be small compared with
the minority carrier diffusion length. If the base width is small, then the minority
carrier electron concentration is a function of both the B-E and B-C junction
voltages.
The current components
The emitter current IE consists of two components. One component of emitter
current IEn is due to the flow of elections injected from the emitter into the base.
The other component IEp is due to the holes injected across the B-E junction into
the emitter. So IE = IEn + iEp = IEn.
The base current IB consists of the current IEp, the recombination current due to the
recombination of majority carrier holes in the base with electrons injected from the
emitter.
The collector current IB consist of the electrons reaching the B-C junction and swept
to the collector. IC = α IE where α is the common base current gain.
IE = IB + IC
The modes of operation of the transistor
Mode E-B junction B-C junction Application
Forward active Forward-biased Reverse-biased Amplifier
Saturation Forward-biased Forward-biased Switch
Cutoff Reverse-biased Reverse-biased Switch
Reverse active Reverse-biased Forward-biased Rarely used
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Solution to Question 5 (18 points)
Choose the correct answer justifying your choice.
(1) The temperature of an electric heating element is 150°C. At what wavelength
does the radiation emitted from the heating element reach its peak?
(a) 8.9 μm (b)9.8 μm (c) 6.9 μm (d) not stated
λm = C/T = 2.9x10 -3 /423 = 6.9x10 -6 m = 6.9 μ
(2) Electrons are ejected from the metallic surface in a photoelectric tube with
speeds of up to 4.60x10 5 m/s when light with a wavelength of 625 nm is used.
What is the work function of the metallic surface?
(a) 1.38 eV (b) 3.31 eV (c) 2.46 eV (d) not stated
Ф = hc/λ - ½ mv 2 = (6.63x10 -34 )(3x10 8 )/(625x10 -9 ) - ½ (9.1x10 -31 )(4.6x10 5 ) 2
= 3.1824x10 -19 – 9.278x10 -20 = 2.2546x10 -19 J= 1.41 eV
(3) What is the energy of the characteristic x-ray emitted from a tungsten target
when an electron drops from an M shell to a vacancy in the K shell? The atomic
number for tungsten is Z = 74.

(a) 74 keV (b) 68 keV (c) 87 keV (d) not stated
hf = EM – EK = [- (74 – 9) 2 x 13.6]/9 – [(74) 2 (-13.6/1)] = 68 keV
(4) Which is the following important in the operation of the laser …….
(a)spontaneous emission (b)stimulated emission (c)absorption (d) all what stated
When light is incident on a collection of atoms, a net absorption of energy usually
occurs because when the system is in thermal equilibrium, many more atoms are in
the ground state than in excited states. If the situation can be inverted so that
more atoms are in an excited state than in the ground state, however, a net
emission of photons can result. Such a condition is called population inversion.
Suppose an atom is in the excited E2 and a photon with energy hf = E2 – E1 is
incident on it. The incoming photon can stimulate the excited atom to return to the
ground state and thereby emit a second photon having the same energy hf and
traveling in the same direction. The incident photon is not absorbed, so after the
stimulated emission, there are two identical photons: the incident photon and the
emitted photon. The emitted photon is in phase with the incident photon. These
photons can stimulate other atoms to emit photons in a chain of similar processes.
The many photons produced in this fashion are the source of the intense, coherent
light in a laser.
The first photon that triggers the operation of laser is from the spontaneous
emission process. Once an atom is in an excited state, the atom can make a
transition back to a lower energy level, emitting a photon in the process. This
process is known as spontaneous emission because it happens naturally.
(5) The lattice constant of silicon is 5.43 A. Calculate the volume density of silicon
atoms.
(a) 5x10 22 cm -3 (b) 8x10 22 cm -3 (c) 5x10 25 cm -3 (d) not stated
N = 8/(5.43x10 -8 ) 3 = 5x10 22 cm -3

(6) The figure shows the E – k relation. Which material is A and which is B?
(a) Both A and B are semiconductors (b) Both A and B are metals
(c) A is metal and B is semiconductor(d) A is semiconductor and B is metal
Curve A: is the free electron model as E ~ k 2 (E = ℏk 2 /2m = p 2 /2m). So curve A
represents E-k relation for a metal.
Curve B: there is a curvature at the zone boundary with slope zero as a result of
motion of electron in periodic crystal. So curve B represent s E-k relation for a
semiconductor.
(7) Calculate the thermal equilibrium hole concentration in intrinsic silicon at T =
400 K. Assume that the Fermi energy is 0.27 eV above the valence band energy.
For silicon at 300 K, NV = 1.04x10 19 cm -3 .
(a) 4.43x10 15 cm -3 (b) 6.43x10 17 cm -3 (c) 6.43x10 15 cm -3 (d) not
stated
kT(400) = kT(300)x(400/300) = (0.026)(400/300) =0.03467 eV = 0.0347 eV
NV(400) = NV(300)x(400/300) 3/2 = 1.04x10 19 x(400/300) 3/2 = 1.6x10 19 cm -3
po = NV e -(EF – EV)/kT = 1.6x10 19 e -0.27/0.0347 = 6.68x10 15 cm -3
(8) Assume that, in an n-type gallium arsenide semiconductor at T = 300 K, the
electron concentration varies linearly from 1x10 18 cm -3 to 7x10 17 cm -3 over a
distance of 0.1 cm. What is the diffusion current density if the electron diffusion
coefficient is Dn = 225 cm 2 /s.
(a) 108 A/cm 2 (b) 808 A/cm 2 (c) 618 A/cm 2 (d) 708 A/cm 2
Jn = eDn dn/dx = (1.6x10 -19 18 17
1x10
7x10
2
)(225)(
) 108A
/ cm
0.
1
(9) The Zener breakdown that occurs in reverse-biased pn junction diode is due to
(a) the forward-bias voltage. (b) the tunneling of electrons. (c) the built-in
voltage. (d) the diode capacitance.
Zener breakdown occurs in highly oped pn junctions through a tunneling
mechanism. In a highly doped junction, the conduction and valence bands on
opposite sides of the junction are sufficiently close during reverse bias that
electrons may tunnel directly from the valence band on the p side into the
conduction band on the n side.

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