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ENE 5330 Midterm Exam
Fall 2017
Name: Student number:
- (20%) (a) How many atoms are there in the unit cell characterizing the Si lattice?
(b) What is the density of atoms per cm
- in the Si lattice at room temperature? The lattice constant for Si is
5.43 angstrom at room temperature.
(c) Determine the center-to-center distance between the nearest neighbors in the Si lattice
(d) Calculate the density of Si at room temperature. The atomic weight of Si is 28.086 g/mol. The number of
atoms in a mole is called Avogadro's number, the value of which is approximately 6.022 ร 10
23 .
(e) Calculate the (100) and (110) surface density (# of atoms per area).
Solution
(a)
Si lattice is diamond structure, hence: there are three type of atoms in the unit cell
8 corner atoms ร
1
8
= 1 atom
6 face atoms ร
1
2
= 3 atoms
4 enclosed atoms ร 1 = 4 atoms
Total of 8 atoms per unit cell.
(b)
Number of density in Si ร
8
๏ฟฝ5.43ร
โ 8 ๐๐๏ฟฝ
3 = 5 ร 10
22
โ 3 .
(c)
From 1.(b), lattice constant for Si is 5.43ร
.
Space diagonal of the unit cell = โ3a = 8r โ r =
โ
8
Center to center distance between the nearest neighbors is
2r = 2 ร
โ
8
๐ = 2 ร
โ
8
ร ๏ฟฝ5.43ร
๏ฟฝ = 2.35ร
.
(d)
Density of Si is
๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐ ๐๐ ๐๐
๐ด๐ด๐๐ด๐ด๐๐๐ โฒ๐ ๐๐๐๐๐๐
ร ๐ค๐ค๐ค๐คโ๐ก ๐๐ ๐๐ค
5 ร 10
22
โ
6.022 ร 10
23
โ
ร 28.086 ๐ค โ ๐๐๐
โ
3
(e)
Surface density of (100) is
4 ร
1
4
( 5. 43 ร 10 โ8^ ๐๐)^2
= 6.78 ร 10
14
โ
Surface density of (110) is
4 ร
1
4
+2ร
1
2
- 43 ร 10 โ8^ ๐๐ ร โ2ร 5. 43 ร 10 โ8^ ๐๐
= 9.59 ร 10
14
โ
- (20%) Consider the n=1 eigenfunction (ground state wave function) of an infinite 1D potential well of
width a. The potential energy U(x) is zero inside the well and infinity outside the well.
a) Show that the normalized wavefunction is
1
( ) x sin( x / a ) a
b) Compute < x > and provide an interpretation.
c) Compute < px > and provide an interpretation.
d) Compute
2 < p x >
e) Show that
2 2 2
2 1 0 0
p x
m m a
ฯ ฮต
, whereฮต 1 is the ground state energy.
Solution:
a) ฯ (^) ( x (^) ) = A sin (^) ( kx (^) ) + B cos( kx )
ฯ (^) ( 0 )= 0 โ B = 0
( ) 0
n a k a
( ) sin^
n x A x a
ฯ ฯ
( )
2
0
a x dx A a
โซ
( )
sin
n x x a a
1 (^ )
x sin x a a
b) (^) ( ) ( )
sin 2
a a (^) a x x x x dx x x dx a a
ฯ ฯ ฯ
โซ โซ
The expectation value of position is
2
a , which is the middle of the infinite potential well
- (20%) Consider a two dimensional semiconductor with width W and length L. Assuming the dimensions
are much larger than atomic dimension, derive an expression for the density of states (g (^) 2D) in energy.
Assume a valley degeneracy of gV.
a) Assume a parabolic dispersion near k = 0
2 2
2
C
k E E m
b) Assume a linear dispersion near k = 0
E = E C + > ฯ
Fk
Show details of your work.
Solution:
a)
1 1
2 2
States between &
E E E
k dk k A spin kdk
L W
= ร =
States per unit energy per area @ (^1 )
V D
g dk E g k
ฯ dE
2 2
C C C
k m E E dk m E E k m dE E E
โ = + โ = โ = โ
(^2 )
V V D
g dk g m g k
ฯ dE ฯ
b)
C^1
C F F F
E E dk E E k k dE
(^2 2 )
2 for positive and negative
V V C D F
g dk g E E g k ฯ dE ฯ ฯ
k
- (10%) Consider germanium with an acceptor concentration of Na=
15 cm
- . Consider temperature of 200,
400, and 600K. Calculate the hole concentration and the position of the Fermi level with respect to the
intrinsic Fermi level at these temperatures. Assume that for germanium, NC =1.03x
19 cm
NV=5.35x
18 cm
- , and Eg =0.663 eV are constants.
Solution:
2 = ๐๐ ๐๐ด๐ค๐๐ ๏ฟฝ
and ๐ 0 =
๐๐ โ๐๐
2
๐๐ โ๐๐
2
2
2 , ๐๐ด โซ ๐๐ โ ๐ 0 =
๐๐
2
๐๐
2
2
2
๐ = 200 ๐พ, ๐๐ = 0.0259 ๐ค๐ ร
๐๐ = 3.538 ร 10
10 ๐๐
โ , ๐ 0 = 1 ร 10
15 ๐๐
โ , ๐ธ๐น๐ โ ๐ธ๐น = 0.177 ๐ค๐
๐ = 400 ๐พ, ๐๐ = 0.0259 ๐ค๐ ร
๐๐ = 4.984 ร 10
14 ๐๐
โ , ๐ 0 = 1.206 ร 10
15 ๐๐
โ , ๐ธ๐น๐ โ ๐ธ๐น = 0.0305 ๐ค๐
๐ = 600 ๐พ, ๐๐ = 0.0259 ๐ค๐ ร
๐๐ = 1.234 ร 10
16 ๐๐
โ , ๐ 0 = 1.285 ร 10
16 ๐๐
โ , ๐ธ๐น๐ โ ๐ธ๐น = 0.00210 ๐ค๐
7 (10%) The Fermi level in n-type germanium at T=300K is 245 meV below the conduction band and 200
meV below the donor level with a donor concentration of 10
18 cm
- . Determine the probability of finding an
electron (a) in the donor level and (b) in the conduction band minimum. Also calculate electron density n 0.
Solution:
(a)
For the donor level
= 8.852 ร 10
โ
(b)
We have
= 7.794 ร 10
โ
๏ฟฝ = 1.03 ร 10
19 ๐๐
โ ร 7.794 ร 10
โ = 8.028 ร 10
14 ๐๐
โ
