Download solid state 7th chapter 1 and more Lecture notes Physics in PDF only on Docsity!
Anderson & Anderson 1 2/15/
SOLUTIONS CHAPTER 1
1.1. Show that Equation (1.6) follows from Equation (1.3).
Solution: Equation (1.3) is dEP =
q^2
4 πε 0 r^2
dr. Integrating both sides we obtain
dEP = q^2
4 πε 0 r
2 dr^ =^
q^2
r ∫ ∫ ∫^2 dr^ =
q^2
r
⎣⎢^
To find the constant, we employ the boundary condition that at r = ∞ , EP=Evac : 2
0
P vac 4
q E r E const const
∴ const = Evac and 2
(^40) P vac vac
q E E E
, Equation (1.6)
1.2. Consider a lithium nucleus, of charge +3q. Calculate the first three electron energies for an electron in a Li ++^ ion, using the Bohr model. We repeat the analysis that we used for the hydrogen atom, except that now the charge of the nucleus Q 1 is equal to 3 q = +3(1.6 × 10 −^19 ) C. The results of the key steps are 2 1 2 2 2 0 0
Q Q q F
πε r πε r
2
0
P vac 4
q E r E
πε r
2 2 2 0
mv q
r πε r
mv r n n = n h
2
0
n 4
q v
πε n
h⎝ ⎠
2
n 3 r n mq
h
2 4 2 2 2 0
K
m q E
πε n
h ⎝^ ⎠
4 2 2 2 0
n Pn Kn vac
mq E E E E
πε n
h ⎝^ ⎠
Thus E 1 = Evac − 9 13.6( eV )= Evac − 122 eV
Anderson & Anderson 2 2/15/
and
(^2 )
vac vac
eV E = E − = E − eV
and
(^3 )
vac vac
eV E = E − = E − eV
1.3. Show that Equations (1.12) and (1.13) follow from (1.8) and (1.11).
Equation (1.8) is:
2 2 2 0
mv q
r πε r
Multiply both sides by r 2 and divide by v :
2
(^40)
q mvr
πε v
= which from Equation (1.11) is
2
(^40)
q mvr n
πε v
= = h
Solving the right hand equality for v :
2
(^40)
q v
πε n
h
(Equation (1.13))
Solving the left hand equality for r substituting in v :
2 2
n n^^4 n^^4 n r mv mq mq
h h^ h^ h (Equation (1.12)).
1.3. In each of the potential energy distributions in Figure 1P.1, sketch the magnitude and direction of the force on the electron.
Anderson & Anderson 4 2/15/
EP ( C ) = Etotal − 10 eV. Therefore, the kinetic energy of the electron is
EK ( C ) = Etotal − E P ( C ) = Etotal − ( E total − 10 eV ) = 10 eV , and
( ) (^ )^196
31
v E^ K C eV^ eV^ J m s m kg
− −
⋅ ×
= = = ×
×
There is no force on the electron because the slope of the potential energy is zero, but the electron continues moving to the right because it still has kinetic energy. At Point D, the electron has made a collision. Its total energy is now 10eV less than it was- the extra 10 eV of energy was transferred to the atom it collided with. The potential energy is EP ( D ) = Etotal − 10 eV. The electron’s total energy is now Etotal ( new ) = Etotal ( original ) − 10 eV , which is the same as its potential energy. Therefore, the kinetic energy of the electron is zero, and so is the velocity. There is no force on the electron because the slope of the potential energy is zero.
1.6. Find the kinetic energies in each of the following. Express all your answers in electron volts. (a) an electron in the first allowed energy states of the hydrogen atom (according to the Bohr model, Eq. (1.14))
We use the relation
4 (^2 ) 0
K
mq E
πε n
h
(^4 ) 2 2 2 2 2 2 2
K
E kg E C E E F m E J s E J eV
(b) a free electron, initially at rest at the back of a cathode ray tube in your television, accelerated through a potential of 10 kV to strike the phosphor layer An electron accelerated through 10kV acquires an energy of 10KeV. Since this electron was initially at rest, its kinetic energy is now 10 KeV, or 10,000 × 1.6 × 10 −^19 J = 1.6 × 10 −^15 J. (c) a tiny, drifting dust particle, of mass 1 μ g and velocity a leisurely 1 mm/s.
We use 2 9 3 2 16
K 2 2
E = mv = ⋅ −^ kg ⋅ −^ m s = × − J = keV
Anderson & Anderson 5 2/15/
1.7. For the following semiconductor materials, indicate to what degree you expect covalent or ionic bonding, and why: Ge GaP InGaAsP HgCdTe
Ge: This element is in Column IV of the periodic table. Each atom in the crystal shares 4 electrons with each neighboring atom, and in return “receives” 4 electrons. Because every atom is “giving” and “receiving” the same number of electrons, the charge is evenly distributed between atoms, and the bonding is covalent. GaP: Gallium is in Column III and phosphorous is in Column V. Thus in the bonding, indium has less positive charge than the phosphorus, so the electrons spends slightly more time near the P atoms than the Ga. This bonding is largely covalaent but with a slightly ionic flavor. InGaAsP: Indium and gallium are in Column III and arsenic and phosphorous are in Column V, so the bonding has the same character as that of InP, mostly covalent but somewhat ionic. HgCdTe: Mercury is in Column II, while cadmium and tellurium are in column VI. This II-VI semiconductor material has a largely ionic but slightly covalent bonding. The electrons spend quite a bit of their time nearer the column VI atoms because of those atoms’ greater positive charge.
1.8. For each of the semiconductors below, draw (to scale) the energy band diagrams:
Si: Eg =1.12 eV. χ =4.05 eV
Ge: Eg =0.67 eV, χ =4.0 eV
GaAs: Eg =1.43 eV, χ =4.07 eV
InP: Eg =1.35 eV, χ =4.35 eV
Anderson & Anderson 7 2/15/
ψ ( x ) = Ae
jx^2 m^0 ( E^ − E^0 ) h^2
⎛ ⎝
⎜⎜ ⎞ ⎠
⎟⎟
− jx^2 m^0 ( E^ − E^0 ) h^2
⎛ ⎝
⎜⎜ ⎞ ⎠
⎟⎟
We wish to show this is a solution to Equation (1.37): d^2 ψ ( x ) dx^2
2 m 0 h^2
( E − E 0 )ψ ( ) x = 0. We begin by finding the derivatives:
d ψ ( x ) dx
= A j
2 m 0 ( E − E 0 )
h^2
e
jx^2 m^0 ( h E^2 − E^0 )
2 m 0 ( E − E 0 )
h^2
e
− jx^2 m^0 ( h E^2 −^ E^0 )
and
d^2 ψ ( x )
dx^2
= A j
2 m 0 ( E − E 0 )
h^2
2 e
jx^2 m^0 ( h E^2 −^ E^0 )
2 m 0 ( E − E 0 )
h^2
2 e
− jx^2 m^0 ( h E^2 − E^0 )
= − A
2 m 0 ( E − E 0 )
h^2
⎟ e^
jx^2 m^0 ( h E^2^ − E^^0 )
2 m 0 ( E − E 0 )
h^2
⎟ e^
− jx^2 m^0 ( h E^2 −^ E^0 )
= Ae
jx^2 m^0 ( h E^2 − E^0 )
⎧ − jx^2 m^0 ( h E^2^ −^ E^0 ) ⎨
2 m 0 ( E − E 0 )
h^2
Substituting into Equation (1.37) we obtain
Ae
jx^2 m^0 ( h E^2^ −^ E^0 )
⎛ − jx^2 m^0 ( h E^2 −^ E^0 )
⎝
⎟⎟ −^
2 m 0 ( E^ −^ E 0 )
h^2
2 m 0 ( E − E 0 )
h^2
Ae
jx^2 m^0 ( h E^2 −^ E^0 )
⎛ − jx^2 m 0 ( h E^2^ −^ E 0 )
⎝
⎟⎟ =^0
From Equation (1.40), K can be taken as either the positive or negative root. If we choose the opposite sign, then from Equation (1.38),
ψ ( x ) = Ae jKx^ + Be −^ jKx^ , it would be the same as changing the sign of x. Thus
the sign of K reverses the direction of propagation of the wave.
1.14. (a) Calculate the de Broglie wavelength of i) a free electron with 1 eV of kinetic energy
2 1 1.6 10 19 K 2
p E eV J m
= = × − =
sec
kg m p −^ kg −^ J −^
= × × = ×
Anderson & Anderson 8 2/15/
λ =
h p
6.63 × 10 −^34 J − s 539 × 10 −^27 kg ⋅ m / s
= 1.23 nm
b) an electron with 10 keV of kinetic energy
19 1 104 1.6^10 1.6 1015 K
J
E eV J eV
− = × ⎛^ × ⎞= × − ⎜ ⎟ ⎝ ⎠
p mEK kg J kg m s = = × −^ × −^ = × − ⋅
λ = 0.0123 nm
iii) a tiny, drifting dust particle of mass 1 μ g and a leisurely velocity of 1 mm/s
9 3 12 22
m g kg v mm s m s p mv kg m s h p m
− − − −
= = ×
iv) yourself, walking at 4 mph on your way to class..
self, about 70 kg, v =4mph=1.8m/s
p=mv =70 kg×1.8m/s=126 kg-m/s
λ=h/p=5.26× 10 -^32 m
(b)What is a typical size of an atom? You begin to see why quantum mechanics and the wave description are not useful for large objects. An atom is on the order of 10-10^ m. Even the 10keV electron's wavelength is a 10th of that, and the dust particle's wavelength is ridiculously small.
Anderson & Anderson 10 2/15/
K (^) x
E
π a
2 π a
3 π a
4 π a
5 π a
5 π a
4 π a
3 π a
2 π a
π a
first Brillouin zone
c) Sketch the corresponding v g - K diagram.
The group velocity
g
dE v dK
h
is proportional to the slope of E :
(^)
π (^)
π
π
π
π
π
π
π
π
π
d) In what regions of the E-K diagram are electrons most likely to be found for this material? Electrons will be at the lowest energies in general, so for this material they will be mostly found near the edges of the Brillouin zones, but not at K=.
π
(^)
π
π
π
π
π
π
π
π
π
1.17. Explain the analogy between using a conducting layer thicker than the skin depth to shield a region of space from electromagnetic waves, and the ability of an electron to penetrate a potential barrier.
Although there cannot be an electric field in a bulk metal, electromagnetic radiation will penetrate a short distance into the metal (skin depth). To prevent the radiation from penetrating the metal and leaking through to the other side, the metal should be thicker than the skin depth by some safety factor.
Anderson & Anderson 11 2/15/
Similarly, an electron wave cannot exist in a bulk material whose potential energy is higher than that of the electron, but the electron can penetrate a short distance into this forbidden region. If the barrier is thin enough, the electron can leak through to the other side.
1.18. The infinitely thick potential barrier of Figure 1.18 (a) can be considered a crude approximation to the potential barrier at the surface of a semiconductor (see Figure 1.12). (a) How, then, might you construct a thin potential barrier like that in Figure 1.18 (b)? Thin potential barriers are used in a wide variety of semiconductor devices, including tunnel diodes, contacts, and field effect transistors. One could place two pieces of semiconductor close together with a narrow air gap (another insulator will do) between them.
(b) How would you construct a potential well (thin region of lower potential energy bounded by region of higher potential energy? Potential wells are widely used in lasers, photodectors, and heterojunction bipolar transistors. In this case one would use a thin layer of semiconductor with air on either side.
1.19. a) From the Bohr model, what emission wavelength would you expect for a transition in hydrogen from E2 to E1****? Transitions ending at E1 are
collectively called the Lyman series, and are generally found in the ultraviolet region of the spectrum.
E 2 − E 1 = Evac − 3.4 − ( E vac − 13.6) = 10.2 eV
The wavelength of this transition is
34 8 19
hc J^ s^ m^ s nm E (^) eV J eV
− −
× − ×
×
b) What emission wavelength would you expect from a transition from E4 to
E3****? This is the first emission line in the Paschen series, and is in the infrared. E 4 − E 3 = Evac − 0.85 − ( Evac − 1. 51) = 0.66 eV
,
and λ =
hc E
= 1.9 μ m
1.20. What wavelength of light should you shine on hydrogen to cause electrons to go from E 1 to E 2 by optical absorption?
Anderson & Anderson 13 2/15/
In the 110 plane, the atoms appear arranged in rectangles that are a by a a ⋅ 2 The area density for the (110) plane is 1/ a^2 . For the (111) plane, the atoms are arranged in regular pattern, the fundamental element of which is an isosceles triangle 2 a on a side.
The area of a triangle is
a a bh = a = (^). But, each triangle
contains 1/6 of at atom at each point, so the area density is (^22)
a^ a
c) What if the lattice is FCC instead (still with a =0.5 nm)? Now how many atoms per unit volume are there? The volume of the unit cell is still the volume of the cube, but now the number of atoms per unit cell is different. There are still 8 corner atoms, each contributing 1/8 of an atom to the unit cell, but there are also 6 face atoms. Each of those face atoms is half in a given cell, and half in an adjacent one. Thus the number of atoms/unit cell is 8×1/8 + 6×1/2= atoms per unit cell. The number of atoms per unit volume is thus 4 atoms/1.25× 10 -28^ m 3 =32× 1027 m -^.
