Chapter 1 problems 1.1 Show that Equation (1.6) follows from Equation (1.3). q2 dr Equation 1.3: dEP = dEP ( r ) = -Fdr = 4pe 0 r 2 We know that EP ( r = ¥) = Evac (Equation 1.4) Substituting in: ò Evac EP dE P = ò ¥ r q2 dr (Equation 1.5) 4pe 0 r 2 Carrying out the integrals: q2 dr ¥ dr Evac - EP = r r2 4pe 0 ¥ ¥ q2 ¥ 1 ¥ = ¥- ¥ 4pe 0 ¥ r r ¥ =- q2 ¥ 1 ¥ q2 0 = + 4pe 0 ¥ 4pe 0 r ¥ r¥ ¥ EP = Evac - q2 4pe 0 r 1.2 Fill in the steps to derive Equations (1.12) and (1.13) from Equations (1.8) and (1.11.) (This is corrected) mv2 q2 Equation 1.8: =0 r 4pe 0 r 2 Equation 1.11: Taking the first and multiplying by rn2 : mvn 2 2 q2 rn rn 2 0 rn 4 0 rn 2 mvn 2 r q2 4 0 mvn rn vn q2 4 0 n vn vn q2 4 0 q2 4 0 n That is Equation 1.13. Now substitute in this expression for vn in Equation 1.8: mvn2 q2 0 rn 4 0 rn2 2 m q2 q2 rn 4 0 n 4 0 rn2 m q2 1 rn 4 0 n 4 0 n 2 2 rn 2 mq which is Equation 1.12. 1.3 Using the Bohr model, find the first three energy levels for a He+ ion, which consists of two protons in the nucleus with a single electron orbiting it. What are the radii of the first three orbits? We repeat the analysis that we used for the Hydrogen atom, except that the charge on the nucleus is nowQ1=+2q. F= Q1Q2 -2q2 = 4pe 0 r 2 4pe 0 r 2 EP (r ) = Evac - 2q2 4pe 0 r mv2 2q2 =0 r 4pe 0 r 2 Thus E1 = Evac - 4(13.6eV) = Evac - 54.4eV and E2 = Evac - 4(13.6eV) / 22 = Evac -13.6eV and E3 = Evac - 4(13.6eV) / 32 = Evac - 6.04eV 1.4. For each of the potential energy distributions in Figure 1P.1, sketch the force on an electron in this region of space, paying attention to magnitude and sign. We know that force is minus the gradient of the potential F = -dEP / dr , so we need to sketch the negative slope. 1.5) Consider the electron in the energy diagram in Figure 1P.2. It is initially at point A , and moving to the left. At point C, it collides with an atom and loses some energy. a) What is its initial potential energy? The potential energy in this region is E0. b) What is its initial kinetic energy? The total energy is 0.1 eV above the potential energy at that point, so the difference, the kinetic energy, is 0.1 eV. c) What is its initial acceleration? The slope of EP =0 in this region, so there is no force and thus no acceleration. d) At point B, what is its potential energy? Kinetic energy? At B the potential energy EP has decreased to about E0-0.1 eV. The total energy has remained the same by conservation of energy, or Etotal=E0+0.1eV. The difference between the total energy and the potential energy is the kinetic energy, or EK=E0+0.1 –(E0-01)=0.2 eV. e) What are its potential and kinetic energies just before the collision? Just before the collision, the potential energy is about E0-0.35 eV, and the kinetic energy is therefore EK=Etotal-EP=E0+01-(E0-0.35)=045 eV. f) What are its potential and kinetic energies just after the collision? After the collision, the electron has lost some energy. Its total energy is now E0-0.2 eV, the potential energy is still E0-0.35 eV , so the kinetic energy is +0.15 eV. g) Is the electron still moving after the collision? It still has kinetic energy so it has to be moving. h) Where did the energy lost by the electron go? It was given to the atom that it hit. g) What is the direction of force on the electron after the collision? The force is minus the gradient of the potential. The slope dEp/dr is positive here, so the force is in the negative r direction, or to the left. h) What is the direction of travel of the electron after the collision? There is no way to know that from the figure. We know the direction of force but the electron might be accelerating or decelerating, depending on which was it is going. 1.6. Find the kinetic energy of each of the following. Express all your answers in electron volts. a. An electron in the lowest energy level of a hydrogen atom according to the Bohr model b. A grain of pollen weighing 10 ng and drifting at 1 mm/s 2 æ gæ m æ 2æ 10E 9 g 1E 3 1mm/ s ( ) ( ) æ æ æ1000mmæ æ æ æ kg æ mv2 EK = = 2 2 (1E - 12 ) = 5E - 18J = 2 2 (1000 ) æ 1eV æ Converting Joules to eV: 5E -18J æ = 31eV æ1.6E -19J æ æ c. An ant weighing 3 mg and running at 1 cm/s 2 æ gæ m æ 2æ ( 3E - 3 ) gææ1E - 3 kg ææ(1cm/ s) ææ100cmææ mv2 EK = = 2 2 ( 3E - 6 ) = 1.5E - 10J = 2 2 (100 ) æ 1eV æ Converting Joules to eV: 1.5E -10J æ = 9.3 æ10 8 eV æ1.6E -19J æ æ 1.7 For each of the semiconductors below, indicate to what degree you expect covalent or ionic bonding, and why: Silicon 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. Gallium nitride Gallium is in Column III and nitrogen is in Column V. Thus in the bonding, gallium has less positive charge than the nitrogen, so the electrons spends slightly more time near the N atoms than the Ga. This bonding is largely covalent but with a slightly ionic flavor. Silicon carbide Both silicon and carbon are in Column IV of the periodic table, so we expect covalent bonding. Cadmium telluride Cadmium is in Column II and tellurium is in Column IV, so the bonds are expected to be more ionic than covalent. 1.8. For each of the following semiconductors, sketch (to scale) the energy band diagrams: a) Indium phosphide Eg=1.35 eV, χ=4.4 eV b) Germanium Eg=0.67 eV, χ=4.0 eV c) Gallium nitride: Eg=3.437 eV, χ=4.1 eV d) Amorphous silicon dioxide Eg=9 eV, χ=0.9 eV 1.9. What minimum energy must an electron at the bottom of the conduction band in gallium phosphide gain to become free of the crystal? Repeat for an electron at the top of the valence band. The energy needed to move an electron from the bottom of the conduction band to become free of the crystal is the electron affinity. From Table 1.1 that value is χ=4.3 eV. To become free of the crystal from the top of the valence band, the electron need to gain an energy equal to χ+Eg=4.3+2.25=6.45 eV. 1.10. A non-degenerate semiconductor cannot conduct current at absolute zero (degeneracy will be discussed in Chapter 2). How much energy must at least one electron obtain in germanium before conduction is possible? At least one electron must be excited into the conduction band, so it must gain an energy equal to the band gap for germanium of 0.67 eV. 1.11. At room temperature in a cubic centimeter of Si there will be about 10 billion electrons in the conduction band. a. How many holes are in the valence band? For every electron that is excited to the conduction band, there is a hole left behind, so there are 10 billions holes. b. If electrons are constantly seeking lower energies and recombining with holes (empty states at lower energies), then how can the number 10 billion remain constant? Because the material is at room temperature, electrons are also being excited into the conduction band by thermal energy. The rate of generation and recombination are equal, leading to a constant electron density. 1.12. Suppose the electron in Figure 1.12 is traveling to the right at a constant energy. What happens to it when it approaches the surface of the material? Explain your answer, using the energy diagram. Let us assume the electron is travelling to the left initially. It travels at constant velocity(because Etotal is constant and EP is constant, so EK is constant), until it gets near the surface. Then the potential energy increases (starting at point B). The total energy is constant, so the kinetic energy has to decrease. At point C the total energy is equal to the potential energy, so the kinetic energy is zero and the electron stops. There is a force to the right, however, because of the gradient in the potential energy. So the electron is accelerated to the right. The net result is the electron is reflected from the barrier. 1.13. Show that Equation (1.38) is a solution to Equation (1.37). What is the significance of the positive and negative values of K? Equation 1.38: y ( x) = AejKx + Be- jKx Equation 1.37: First we find the second derivative: dy ( x) d = AejKx + Be- jKx = AjKejKx - BjKe- jKx dx dx 2 d y ( x) d = AjKejKx - BjKe- jKx = -AK 2 ejKx - (-)BejKx dx2 dx = -K 2 AejKx + Be- jKx ( ) ( ) ( = -K 2y ( x) Substituting in: ) The ψ’s cancel and we have a solution when There are positive and negative roots. When K is positive, the plane wave is travelling to the right, and when K is negative, the wave is traveling to the left. 1.14. a. Calculate the de Broglie wavelength of i. A free electron with 1 eV of kinetic energy h h l= = p mv We find the velocity from the kinetic energy: mv2 æ1.6E - 19J æ EK = = 1eV æ = 1.6E - 19J æ æ 1eV æ 2 2 (1.6E - 19J ) 2EK m = = 5.9E5 m 9.1E - 31kg s m¥ kg - m ¥ p = mv = ( 9.1E - 31kg) ¥5.9E5 ¥ = 5.3E - 25 ¥ s¥ s v= 6.63E - 34 J ¥s ¥kg - m¥ 5.3E - 25 ¥ ¥ s ¥ ¥ kg - m2 One Joule is one 1J = 1 s2 l= h = p æ kg - m2 æ 6.63E - 34 æ æs æ s2 æ æ = 1.25E - 9m = 1.25nm ækg - mæ 5.3E - 25 æ æ s æ æ ii. A free electron with 1keV of kinetic energy mv2 æ1.6E - 19J æ EK = = 1000eV æ = 1.6E - 16J æ æ 1eV æ 2 2 (1.6E - 16J ) 2EK m = = 1.9E7 m 9.1E - 31kg s m¥ kg - m ¥ p = mv = ( 9.1E - 31kg) ¥1.9E7 ¥ = 1.7E - 23 ¥ s¥ s v= ¥ kg - m2 ¥ 6.63E - 34 ¥ ¥s ¥ s2 ¥ ¥ h l= = = 3.9E - 11m = 39 pm p ¥ kg - m¥ 1.7E - 23 ¥ ¥ s ¥ ¥ iii. A grain of pollen weighing 10 ng and drifting at 1 cm/s ¥ g kg ¥¥ cm m ¥ kg - m p = mv = ¥1ng¥ 9 ¥ 1 ¥ = 1E - 14 ¥ ¥ ¥ ¥ 10 ng 1000g ¥¥ s 100cm¥ s ¥ kg - m2 ¥ 6.63E - 34 ¥ ¥s ¥ s2 ¥ ¥ h l= = = 6.63 ¥10 -20 m kg m p ¥ ¥ 1E - 14 ¥ ¥ s ¥ ¥ iv. Yourself, walking at 1.5 m/s on your way to class A 70 kg person walking 1.5 m/s has a momentum of kg - m p = mv = ( 70kg) (1.5m/ s) = 35 s 2 ¥ kg - m ¥ 6.63E - 34 ¥ ¥s ¥ s2 ¥ ¥ h l= = = 1.9 ¥10 -35 m p ¥ kg - m¥ 35 ¥ ¥ s ¥ ¥ b) What is the size of a typical atom? You begin to see why quantum mechanics and the wave description are not useful for large objects. A typical atom might have a diameter on the order of 100 pm. 1.15. What is the wavelength of an electron at the bottom of the E-K relationship of Figure 1.13? What is its kinetic energy there? At the bottom of the E-K diagram, the kinetic energy is zero, therefore the momentum is zero and the wavelength is infinite. 1.16. The laws of classical physics apply to an object in motion if its dimensions are much larger than its deBroglie wavelength. A particle can be considered as a wave (Eq. 1.25), where the wave vector K is K = 2p / l , and for a free particle (constant EP) l = h/ P , where P=mv. Find the wavelength of a car of mass 3000 lbs traveling at 60 miles per hour and of a free electron traveling at 106cm/sec. Note: 2 2 The dimensions of Joules are kg m / s . 3000 lbs=1360 kg 60 mph=27 m/s p = mv = (1360kg) ( 27m/ s) = 3.7E4 kg - m s ¥ kg - m2 ¥ 6.63E - 34 ¥ ¥s ¥ s2 ¥ ¥ h l= = = 1.8E - 39m p ¥ kg - m¥ 3.7E4 ¥ ¥ s ¥ ¥ 1.17 Find the energy of a photon having wavelengths of (a) 1 cm 1.24 1.24 ¥ 1eV ¥ E ( eV ) = = = 5E - 23J ¥ = 0.124 meV ¥1.6E -19J ¥ ¥ l ( m m) ¥1E6 m m¥ 1E - 2m¥ ¥ m ¥ ¥ This wavelength corresponds to the high-energy end of the microwave spectrum. (b) 1 µm 1.24 1.24 = = 1eV l (m m) 1µm This wavelength is in the near infrared. E ( eV ) = (c) 1 nm. 1.24 1.24 E ( eV ) = = = 1keV l (m m) 1E - 3µm This is a soft X-ray photon. 1.18 An AM radio station radiates 100 kW of power at 1 MHz. (a) What is the energy of each photon? ¥ 1eV ¥ E = hn = ( 6.63E - 34J ¥s) 1E6s-1 = 6.63E - 28J ¥ = 1.1E - 9eV ¥6.1E -19J ¥ ¥ ( ) (b) How many photons are radiated per second? E J P = = 100E3 t s If each photon has 6.63×10-28 J of energy, then we are radiating 100 ´ 10 3 J / s photons = 1.5 ´ 10 32 J second 6.63 ´10 28 photon 1.19 An electron is moving at a constant velocity (i.e. free electron) of 105 cm/s. Determine: (a) Its momentum. kg - m æ cmææ 1m æ p = mv = ( 9.1æ10 -31 kg) æ1 ææ = 9.1æ10 -33 æ æ s ææ100cmæ s (b) Its wave vector Since it is a free electron we can treat it classically, so (c) Its de Broglie wavelength kg - m2 h s l= = = 7.2 ´ 10 -2 m kg m p 9.1 ´10 -33 s 6.63 ´10 -34 (d) Its kinetic energy EK = 2 1 2 1 eV æ æ mv = ( 9.1æ10 -31 kg) (10 3 m/ s) = 4.6 æ10 -25 J æ æ = 2.9 æ10 -6 eV -19 æ æ1.6 æ10 J æ 2 2 1.20 Consider the E-K diagram shown in Figure 1P.3. a. Verify that it meets the required criteria: i. E(K) is periodic in K space with period 2π/a. ii. Equivalent extrema exist at K =0,±2 π /a,±4 π /a, . . . . iii. Equivalent extrema exist at K = ± π /a,±3 π /a,±5 π /a, . . . . iv. The slope of the E-K curve is zero at K = 0,± π /a,±2 π /a, . . . . b. Indicate the first Brillouin zone. c. Sketch the corresponding vg-K diagram. d. In what regions of the E-K diagram are electrons most likely to be found for this material? 1.21. Explain the analogy between using a conductor 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. 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. 1.22. The infinitely thick potential barrier of Figure 1.18a 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 1.18b? 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, photodetectors, and heterojunction bipolar transistors. In this case one would use a thin layer of semiconductor with air on either side. 1.23 a. From the Bohr model, what emission wavelength would you expect for a transition in hydrogen from E3 to E1? Transitions ending at E1 are collectively called the Lyman series and are generally found in the ultraviolet region of the spectrum. From Table 1.1, we now that the energies of levels E3 and E1 are Evac-1.51 eV and Evac -13.6 eV respectively so E 3 - E1 = ( Evac - 3.4 ) - ( Evac -13.6 ) = 12.1eV The wavelength of this transition is ( )( ) -34 8 hc 6.63 ´10 J - s 3.0 ´10 m/ s l= = = 103nm E 12.1eV 1.6 ´10-19 J / eV ( ) b. What emission wavelength would you expect from a transition from E3 to E2? This is the first emission line in the Balmer series, and is visible. Again using Table 1.1, we find E2 -Evac=1.51 eV, so E3 - E2 = ( Evac -1.51) - ( Evac - 3.4 ) = 1.9 eV and hc = 653 nm E which appears red. 1.24 a. What wavelength of light should you shine on hydrogen to cause electrons to go from E2 to E4 by optical absorption? From Table 1.1, we know that the energy corresponding to this transition is 2.55 eV, or = 486nm. b. What would happen if you passed a beam of λ = 520 nm (bluish green) through a hydrogen gas? Explain your answer. From Equation 1.74, light of wavelength =430 nm has an energy of 1.24 1.24 EeV 2.88eV . It could not be absorbed, because m 0.43m there is no combination of allowed initial and final states corresponding to this wavelength or energy. Thus, hydrogen is transparent to this radiation. 1.25 In discussing Figure 1.19a, we pointed out that, in a material with a band gap of 2.5 eV, an electron near the top of the valence band could not absorb a photon of energy 2.06 eV, since it would have to end up at a forbidden energy state. a. What about an electron deep in the valence band, more than 2.06 eV below the band edge EV? Why is it unlikely for this electron to absorb the photon? From Equation 1.74, light of wavelength =430 nm has an energy of 1.24 1.24 EeV 2.88eV . It could not be absorbed, because m 0.43m there is no combination of allowed initial and final states corresponding to this wavelength or energy. Thus, hydrogen is transparent to this radiation. b. Why is unlikely for a photon of 2.06 eV to be absorbed by an electron in the conduction band? From Equation 1.74, light of wavelength =430 nm has an energy of 1.24 1.24 EeV 2.88eV . It could not be absorbed, because m 0.43m there is no combination of allowed initial and final states corresponding to this wavelength or energy. Thus, hydrogen is transparent to this radiation. 1.26 For a simple cubic crystalline structure of lattice constant a = 0.40 nm, a. How many atoms are there per unit volume? (Hint: an easy way to proceed is to calculate the volume of the unit cell and the number of atoms per unit cell. Although there are eight atoms involved in any given unit cell, each atom in the simple cubic structure is part of eight different cells, one corner of each. Thus, there are 8 atoms× 1/8/atom per corner or 1 atom per unit cell.) The volume of the unit cell is the volume of the cube, or (0.510-9 m)3, or 1.2510-28 cubic meters. The density of atoms is thus 1/1.25 m3, or 81027 atoms/cubic meter. b. How many atoms per unit area are there in the (100) plane? The (110) plane? The (111) plane? In the (100) plane, we have a grid of atoms on squares a on a side. Each square contains 1/4 of an atom at each corner, or 4 times 1/4 atoms per area a2. Thus the area density of atoms is 1/a2. In the 110 plane, the atoms appear arranged in rectangles that are a by aa 2 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 2a on a side. 1 1a 2 a2 (a 2) The area of a triangle is bh . But, each triangle 2 2 2 2 1 1 6 contains 1/6 of at atom at each point, so the area density is 2 2 . a 3a 2 c. What if the lattice is FCC instead (still with a = 0.40 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 81/8 + 61/2=4 atoms per unit cell. The number of atoms per unit volume is thus 4 atoms/1.2510-28m3=321027 m-3.
