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Unformatted text preview: MatE 153 Spring 2007
Midterm 2
SOLUTIONS
Class average was 75, 79% with extra credit study logs
1. The lone electron in a hydrogen atom is excited to the 4 shell. Calculate the
wavelength of light needed for this excitation. (10 points)
This was looking for an answer exactly like the HW3, Q2b.
The electron is initially in the ground state (n=1). The energy difference between the two
levels is equal to the energy of the light. mq 4
E I = 2 2 = 13.6eV
8ε 0 h
En = − Z2E I
n2 1⎞
⎛1
ΔE = −12 ⋅ 13.6eV⎜ 2 − 2 ⎟ = 12.75eV
⎝4 1 ⎠
hc
ΔE =
λ
m
4.136 × 10 −15 eV ⋅ s ⋅ 3 × 10 8
ΔE
s = 9.7 × 10 −8 m
λ=
=
hc
12.75eV
2. Write the electronic configuration for Vn which has 23 electrons. (5 points)
This was the same as HW3, Q1. This you need to fill in using the chart with the arrows
from lecture or Fig 3.35.
1s22s22p63s23p64s23d3
3. Give the four quantum numbers for the 5 outermost electrons for Vn. (10 points)
This was looking for an answer exactly like the HW3, Q3.
n
l
ml
ms
4
0
0
½
4
0
0
½
3
2
2
½
3
2
1
½
3
2
0
½
4. Why is He2 unlikely? (5 points)
When two He atoms come together, the 1s wavefunction of each atom overlap. They
constructively interfere to form the bonding wavefunction and destructively interfere to
form the antibonding wavefunction. The antibonding wavefunction is significantly
higher in energy due to the modes. For the He2 molecule, there are 2 electrons in the
lower energy bonding orbital and two electrons in the high energy antibonding orbital.
This makes the energy of the He2 molecule significantly higher than 2 lone He atoms. 1 5. What assumptions are different between Boltzmann and FermiDirac probability
functions? (5 points)
See the paragraph on page 313. Boltzmann probability does not account for Pauli’s
exclusion Principle. FermiDirac on the other hand, does take into account that a state
must be empty in order for an electron to move into it.
6. Explain why Si tetrahedrally bonds. (5 points)
See the paragraph on page 374. The angular relationship of the bonds in crystalline silicon
in the shape of a tetrahedral come from the angular relationship of the electron orbitals in a
Si atom. The 3s and 3p levels are very close in energy and interact to form 4 new lobes.
This is known as sp3 hybridization. The lobes are spread out from each as far as possible,
leading to the tetrahedral shape. Each Si atom has 1 electron per lobe. It covalently bonds
with 4 other Si atoms to fill the lobes, 8 total electrons.
7. Using the data in the table below, calculate the number of effective valence electrons
for Ag at 0K. (10 points)
This was just like HW4, Q1.
The most common math error is that folsk left EF in eV. The units won’t work out. In
general, if you see other terms like kg or m, you know you need to convert energy to
Joules. E F0 ⎛ h2
=⎜
⎜8⋅m
e
⎝ ⎞⎛ 3n ⎞
⎟⎜ ⎟
⎟⎝ π
⎠
⎠ 2/3 1.6 × 10 −19 J
J ⋅ s2
−31
5.5eV
⋅ 8 ⋅ 9.1 × 10 kg
eV
electrons
π 2 / 3 E F 0 ⋅ 8m e π 2 / 3
kg ⋅ m 2
n=
= 5.8 × 10 28
=
2
2
3
3
h
m3
6.626 × 10 −34 J ⋅ s ( n= ) # valence ⋅ d ⋅ N A
M At electrons
g
⋅ 107.87
3
n ⋅ M At
electron
mol
m
# valence =
=
= 0.997 ≈ 1
3
d ⋅ NA
atom
g ⎛ 100cm ⎞
23 atoms
10.49 3 ⎜
⎟ ⋅ 6.02 × 10
mol
cm ⎝ m ⎠
8. CdTe is doped with In. The In is a substitutional dopant that replaces Cd. (5 points)
a. Is In an acceptor or donor? In is one column over in the periodic table from
Cd. This means it has one more valence electron relative to Cd, making it a
donor.
5.8 × 10 28 b. Draw a band diagram for this material. Label the important points.
Your figure should look like 5.8b or 5.10. 2 9. Explain, using figures or text what the figure below represents and how it was derived.
(5 points)
See Figure 5,7 and the text describing it. To get full credit, you needed to describe it for
both electrons in the CB and holes in the VB. The top plot is the density of electrons as a
function of energy in the CB. It is derived by multiplying the density of states in the CB
(g(E) from EC to EC + Χ) with the probability of finding an electron (f(E). The bottom plot
is the density of holes as a function of energy in the VB. It is derived by multiplying the
density of states in the CB (g(E) from 0 to EV) with the probability of finding a hole, that is
of not finding an electron (1f(E). 3 ...
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