Physics 107 Problem 10.2 O. A. Pringleweb.mst.edu/~sparlin/phys107/homework/chapter10.pdfThe...
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Transcript of Physics 107 Problem 10.2 O. A. Pringleweb.mst.edu/~sparlin/phys107/homework/chapter10.pdfThe...
Physics 107 Problem 10.2 O. A. Pringle
Show that the first five terms in the series for the Madelung constant of NaCl are
α 612
2
8
3
6
2
24
5
..
To do this problem, you really must draw a picture of the NaCl structure, find the number of nearest, 2nd nearest, and 3rd 4th 5th nearest neighbors of a reference ion. Then you need to find the distance from the reference ion to the nearest neighbors. α is just
αR
near_neighbor_shells
sign( )number_of_neighbors
neighbor_distance.
You will need to draw a picture of the NaCl structure to see the following:
Neighbor Number Distance
nearest 6R
2nd 12 R 12 12.
3rd 8 R 12 12 12.
4th 6 R 22.
5th 24 R 22 12.
There is a WordPerfect file named Prob10-2.wp, which will be downloaded along with this homework file, which contains a 1-dimensional view of the NaCl structure, along with a description of the nearest neighbor locations. Try loading this file into WordPerfect if you are having trouble finding the near-neighbor shells.
Physics 107 Problem 10.3 O. A. Pringle
(a) The ionization energy of potassium is 4.34 eV and the electron affinity of chlorine is 3.61 eV. The Madelung constant for the KCl structure is 1.748 and the distance between ions of opposite sign is 0.314 nm. On the basis of these data only, compute the cohesive energy of KCl.
Key words here are "on the basis of these data only." At first reading, you think Beiser means calculate V using
Vα e2.
4 π. ε 0. r 0
.1
1
n. n 9
If I give you a problem like this, and say something like "use n=9," I mean exactly that.
However, in this case, Beiser wants you to ignore the overlap repulsive term, and use the Coulomb energy only. That will produce a discrepancy between calculated and measured cohesive energy. In part b, you resolve the discrepancy by including the 1-1/n term.
1
So Beiser wants us to do this first part using
Vα e2.
4 π. ε 0. r 0
.
Let's see what we get.
e 1.602 1019. ε 0 8.854 1012. r 0 0.314 109. α 1.748
Vα e2.
4 π. ε 0. r 0
.
V 1.28410 18.=
VV
1.602 1019.
V 8.015= This is the lattice energy.
The cohesive energy is (V.eV-V.ionization+V.affinity)/2.
V ion 4.34 V affin 3.61
E coh V V ion V affin
E coh 7.285= eV This is where Beiser gets -7.29 eV.
Per ion pair,E cohesive
E coh
2
E cohesive 3.643= eV
(b) The observed cohesive energy of KCl is 6.42 eV per ion pair. On the assumption that the difference between this figure and that obtained in a is due to the exclusion-principle repulsion, find the exponent n in the formula Br^-n for the potential energy arising from this source.
V 8.015= per ion pair
E obs 6.42 is the measured value
We need to use
E obsα e2.
4 π. ε 0. r 0
.1
1
n. V ion V affin
or
E obs V 0 V ion V affin
To get V.0, the lattice energy which includes the core repulsion term. Once we have V.0, we can calculate n
2
V 0 E obs V ion V affin This problem is subtle, and a bit confusing. You should study it well. It is not fundamentally difficult; you just have to work carefully.V 0 7.15= eV
Solve V 0 V 11
n. for n:
nV
V 0 V
n 9.262=
Physics 107 Problem 10.4 O. A. Pringle
Repeat Exercise 3 for LiCl, in which the Madelung constant is 1.748, the ion spacing is 0.257 nm, and the observed cohesive energy is 6.8 eV per ion pair. The ionization energy of Li is 5.4 eV.
(a) The ionization energy of lithium is 5.4 eV and the electron affinity of chlorine is 3.61 eV. The Madelung constant for the KCl structure is 1.748 and the distance between ions of opposite sign is 0.257 nm. On the basis of these data only, compute the cohesive energy of Cl.
Key words here are "on the basis of these data only." At first reading, you think Beiser means calculate V using
Vα e2.
4 π. ε 0. r 0
.1
1
n. n 9
If I give you a problem like this, and say something like "use n=9," I mean exactly that.
However, in this case, Beiser wants you to ignore the overlap repulsive term, and use the Coulomb energy only. That will produce a discrepancy between calculated and measured cohesive energy. In part b, you resolve the discrepancy by including the 1-1/n term.
So Beiser wants us to do this first part using
Vα e2.
4 π. ε 0. r 0
.
Let's see what we get.
e 1.602 1019. ε 0 8.854 1012. r 0 0.257 109. α 1.748
Vα e2.
4 π. ε 0. r 0
.
V 1.56910 18.=
VV
1.602 1019.V 9.793= This is the lattice energy.
3
The cohesive energy is (V.eV-V.ionization+V.affinity)/2.
V ion 5.4 V affin 3.61
E coh V V ion V affin
E coh 8.003= eV
Per ion pair, E cohesiveE coh
2
E cohesive 4.002= eV
(b) The observed cohesive energy of LiCl is 6.8 eV per ion pair. On the assumption that the difference between this figure and that obtained in a is due to the exclusion-principle repulsion, find the exponent n in the formula Br^-n for the potential energy arising from this source.
V 9.793= per ion pair
E obs 6.8 is the measured value
We need to use
E obsα e2.
4 π. ε 0. r 0
.1
1
n. V ion V affin
or
E obs V 0 V ion V affin
To get V.0, the lattice energy which includes the core repulsion term. Once we have V.0, we can calculate n
This problem is subtle, and a bit confusing. You should study it well. It is not fundamentally difficult; you just have to work carefully.
V 0 E obs V ion V affin
V 0 8.59= eV
Solve V 0 V 11
n. for n:
nV
V 0 V
n 8.14=
4
Physics 107 Problem 10.10 O. A. Pringle
Gold has an atomic mass of 197 u, a density of 19.3x103 kg/m3, a Fermi energyof 5.54 eV, and a resistivity of 2.04x10-8 Ω. Estimate the mean free path in atom spacings between collisions of the free electrons in gold under the assumption that each gold atom contributes one electron to the electron gas.
The mean free path can be calculated from
ρ2 m. v F
.
N e2. λ.or λ
2 m. v F.
N e2. ρ.
All we need to do is use the given information to calculate N, the free-electron density in gold, and v.F, the Fermi velocity. We plug those values and the constants into the equation to get λ.
Calculate N:N
kg
m3
number_gold_atoms
kg. 1_electron
gold_atom.
N19.3 103.
1.6604 1027. 197.
This agrees with the value listed in my Solid State textbook.N 5.9 1028.=
The Fermi velocity comes from the Fermi energy.
ε F1
2m. v F
2.
v F
2 ε F.
m
Don't forget to multiply ε.F by e if ε.F is in eV.
v F
2 ε F_eV. e.
m
Define the other parameters
m 9.11 1031. Remember, the electrons are the charge carriers.
ε F_eV 5.54e 1.602 1019. ρ 2.04 108.
v F
2 ε F_eV. e.
m
v F 1.396 106.= looks reasonable
5
λm vF
.
N e2. ρ.
λ 4.116 108.= meters
λ nmλ
10 9
λ nm 41.165= nanometers
A good estimate of the atomic spacing is 1 over the cube root of N, the atomic density.
a N
13
a 2.569 1010.=
In units of atomic spacings, the mean free path is
λ aλa
λ a 160.258=
An electron travels an average of 320.5 atomic spacings in gold before it gets scattered.
6
Physics 107 Problem 10.11 O. A. Pringle
Silver has an atomic mass of 108 u, a density of 10.5x103 kg/m3, and a Fermi energy of 5.51 eV. On the assumptions that each silver atom comtributes one electron to the electron gas and that the mean free path of the electrons is 200 atom spacings, estimate the resistivity of silver. (The actual resistivity of silver at 20 C is
1.6x10-8 Ω*m.) The resistivity can be calculated from
ρm vF
.
N e2. λ.
All we need to do is use the given information to calculate N, the free-electron density in gold, and v.F, the Fermi velocity, and λ.
Calculate N:N
kg
m3
number_gold_atoms
kg. 1_electron
gold_atom.
N10.5 103.
1.6604 1027. 108.
N 5.855 1028.= This agrees with the value listed in my Solid State textbook.
The Fermi velocity comes from the Fermi energy.
ε F1
2m. v F
2.
v F
2 ε F.
m
Don't forget to multiply ε.F by e if ε.F is in eV.
v F
2 ε F_eV. e.
m
Define the other parameters
m 9.11 1031. Remember, the electrons are the charge carriers.
ε F_eV 5.51e 1.602 1019.
v F
2 ε F_eV. e.
m
v F 1.392 106.=
We are given that λ is 200 atomic spacings.
A good estimate of the atomic spacing is 1 over the cube root of N, the atomic density.
a N
1
3 a 2.575 1010.=
Thus λ 200 a.
7
λ 5.15 108.= meters
looks reasonable
λ nmλ
10 9
λ nm 51.505= nanometers
Finally, calculate ρ
ρm vF
.
N e2. λ.
ρ 1.639 108.=
8
Physics 107 Problem 10.15 O. A. Pringle
(a) Why are some solids transparent to visible light and others opaque?
Visible light has wavelengths ranging fromλ 1 400 nm to λ 2 700 nm
c 3 108. h 6.63 1034. e 1.6 1019.
The energies range from
E 2h c.
λ 2 10 9. e.
E 2 1.776= eV
to
E 1h c.
λ 1 10 9. e.
E 1 3.108= eV
Solids having band gaps less than 3.1 eV, or having partly filled energy bands wider than 3 eV wide can absorb visible photons, and are opaque. Metals, are a good example of opaque solids.
Solids having a filled band, a band gap greater than 3.1 eV, and an empty band above that are incapable of absorbing visible light, and so are transparent. Glass and diamond are good examples.
Silicon has a 1.17 eV band gap, so that silicon can absorb visible light, and silicon appears metallic.
Many solids that should be transparent (wide band gaps) are opaque because of impurities that scatter light.
(b) The forbidden band is 1.1 eV in silicon and 6 eV in diamond. To what wavelengths of light are these substances transparent.
Silicon is transparent to light having energies less than 1.1 eV, or wavelengths greater than
λh c.
1.1 e.
λ 1.13 106.= or wavelengths greater than 1130 nm
Diamond is transparent to light having energies less than 6 eV, or wavelengths greater than
λh c.
6 e.
λ 2.072 107.=
or wavelengths greater than 207 nm
9
Physics 107 Problem 10.16 O. A. Pringle
The forbidden band is 0.7 eV in germanium and 1.1 eV in silicon. How does the conductivity of germanium compare with that of silicon at (a) very low temperatures and (b) room temperature?
(a) Semiconductors conduct when electrons are excited across the band gap into the conduction band. The excitation, in this case, takes place due to thermal energy. At very low temperatures, there is not enough thermal energy to excite any significant number of electrons across the band gap in either germanium or silicon, so both are insulators at very low temperatures.
(b) At higher temperatures, where thermal energy is available to excite electrons across the band gap, germanium, with its smaller gap of 0.7 eV, will always be a better conductor that silicon, with its gap of 1.1 eV.
It can be shown that the relative number of atoms having electrons excited into the conduction band is proportional to
expE g
k T.=
Because kT at room temperature is about 0.026 eV, the relative number of conduction electrons in silicon is
exp1.1( )
0.0264.227 1019.=
and in germanium is
exp0.7( )
0.0262.03 1012.=
There are 7 orders of magnitude more electrons in germanium's conduction band, but in fact both are poor conductors because even in germanium there are very few electrons in the conduction band.
10
Physics 107 Problem 10.21 O. A. Pringle
Phosphorus is present in a germanium sample. Assume that one of its five valence electrons revolves in a Bohr orbit around each P+ ion in the germanium lattice.
a) If the effective mass of the electron is 0.17 m.e and the dielectric constant of germanium is 16, find the radius of the first Bohr orbit of the electron.
The first step is to figure out what this problem is asking. You are told to consider the electron as orbiting a nucleus of charge +1. The effective mass of the orbiting electron is 0.17*me. From way back in chapter 4 we know how to calculate Bohr radii. The equation to use is 4.13.
r n
n2 h2. ε 0.
π m e. e2.
The only really tricky part is knowing that a dielectric constant of 16 means to use 16*ε0
in this equation.
Define constants and plug numbers into the equation:
n 1 h 6.626 1034. ε 0 8.854 1012. e 1.602 1019. m e 9.11 1031.
ε 16 ε 0. m 0.17 me
.
r 112 h2. ε.
π m. e2.
r 1 4.981 109.= about 5 nm
b) The energy gap between the valence and conduction bands in germanium is 0.65 eV. How does the ionization energy of the above electron compare with this energy and with kbT at room temperature?
The ionization energy of this orbiting electron is given by equation 4.15, using the effective mass, m, n=1 because the electron is in its lowest energy level, and ε instead of ε.0.
Em e4.
8 ε2. h2.
E 1.44710 21.= Joules
In eV, the energy is
E eVE
e
E eV 9.03310 3.= eV
This value of the ionization energy is much less than the 0.65 eV energy gap. kT at room temperature is about 0.025 eV, so the ionization energy is less than kT. At room temperature, there is enough thermal energy to ionize the phosphorus and put the electron into the conduction band.
11
Physics 107 Problem 10.22 O. A. Pringle
Repeat exercise 21 for a silicon sample that contains arsenic.
a) If the effective mass of the electron in silicon is 0.31 m.e and the dielectric constant of silicon is 12, find the radius of the first Bohr orbit of the electron.
The solution proceeds as in problem 10.21. The first step is to figure out what this problem is asking. You are told to consider the electron as orbiting a nucleus of charge +1. The effective mass of the orbiting electron is 0.31*me. From way back in chapter 4 we know how to calculate Bohr radii. The equation to use is again 4.13.
r n
n2 h2. ε 0.
π m e. e2.
The only really tricky part is knowing that a dielectric constant of 12 means to use 12*ε0
in this equation.
Define constants and plug numbers into the equation:
n 1 h 6.626 1034. ε 0 8.854 1012. e 1.602 1019. m e 9.11 1031.
ε 12 ε 0. m 0.31 me
.
r 112 h2. ε.
π m. e2.
r 1 2.049 109.= about 2 nm
b) The energy gap between the valence and conduction bands in silicon is 1.1 eV. How does the ionization energy of the above electron compare with this energy and with kbT at room temperature?
The ionization energy of this orbiting electron is given by equation 4.15, using the effective mass, m, n=1 because the electron is in its lowest energy level, and ε instead of ε.0.
Em e4.
8 ε2. h2.
E 4.69110 21.= Joules
In eV, the energy is
E eVE
e
E eV 0.029= eV
This value of the ionization energy is much less than the 1.1 eV energy gap. kT at room temperature is about 0.025 eV, so the ionization energy about equal to kT. At room temperature, there is enough thermal energy to ionize the arsenic and put the electron into the conduction band.
12