Chapter #11 Solutions - Modern Physics for Scientists and Engineers - Andrew Rex, Stephen T. Thornton - 4th Edition

1. For a certain resistor with a normal (that is, room temperature) resistance of 150 Ω, the values of the constants in Equation (11.4) are found to be A = 4.05, B = 4.22, and K = 4.11, in units such that R will be in ohms and T in kelvin. What is the resistance of this resistor at (a) T = 77 K, (b) T = 20 K, and (c) T = 1 K? ... Get solution

1q. Why is the free-electron model not applicable to semiconductors and insulators? Get solution

2. For a nominal 10-Ω resistor (as described in Figure 11.7) the resistance at various temperatures is as follows: R = 10 Ω at T = 293 K, R = 40 Ω at T = 10 K, and R = 5800 Ω at T = 1 K. Determine the constants A, B, and K in Equation (11.4) for this resistor.... Get solution

2q. How does a semimetal differ from a normal conductor? Get solution

3. Consider the experimental arrangement of Figure 11.9, set up to observe the Hall effect. With the power supplies and meters in this configuration, what will be the sign of the voltage on the voltmeter if the sample is a semiconductor in which the majority of charge carriers are holes? Explain. ... Get solution

3q. Compare the resistivities of conductors, semiconductors, and insulators at room temperature. Get solution

4. Hall effect data in an actual student laboratory was as follows: for a doped indium arsenide strip of thickness 0.15 mm, with a current of 100 mA flowing through the strip and a magnetic field of 50 mT perpendicular to the strip, the measured Hall voltage was 11.5 mV. (a) Use these data to find the density of charge carriers. (b) Find the density of charge carriers using the complete data set shown in the accompanying table. ... Get solution

4q. Why does the addition of impurities to a conductor increase its resistivity, whereas the addition of impurities to a semiconductor generally decreases its resistivity? Get solution

5. A pure lead bar 10 cm long is maintained with one end at T = 300 K and the other at 310 K. The thermoelectric potential difference thus induced across the ends is 12.8 μV. Find the thermoelectric power for lead in this temperature range. (Note: Q varies nonlinearly with temperature, but over this narrow temperature range, you may use a linear approximation.) Get solution

5q. Would you expect carbon resistors to be useful as thermometers at room temperature and above? Explain. Get solution

6. The reference junction of an iron-constantan thermocouple is maintained at 0°C, and the other side is at an unknown temperature. Find the unknown temperature if the other side is 3.03 mV higher in potential than the reference junction (see Table 11.3). ... Get solution

6q. Why are semiconductors referred to as non-ohmic? Get solution

7. Assuming that the potential corresponding to any temperature T in Table 11.3 could be known (say, through a computer model) to at least five significant digits, what maximum uncertainty would be allowed in your voltmeter if you were to use an iron-constantan thermocouple to measure temperatures to within 0.01°C? ... Get solution

7q. Use the size of the energy gap to decide whether the following should tend to be transparent to visible light: conductors, semiconductors, and insulators. Get solution

8. What kind (p-type or n-type) of semiconductor is made if pure germanium is doped with a small amount of (a) phosphorous? (b) gallium? Get solution

8q. Repeat Question 7 considering near-infrared light with λ= 1 m. Get solution

9. Assume a temperature of 300 K and find the wavelength of the photon necessary to cause an electron to jump from the valence to the conduction band in (a) germanium, (b) silicon, (c) InAs, and (d) ZnS. Get solution

9q. A semiconductor has an energy gap Eg. Explain what happens when the semiconductor is bombarded with electromagnetic radiation with wavelength λ hc/Eg . Repeat for λ > hc/Eg . Get solution

10. When an electron in the compound semiconductor AlAs makes a transition from the conduction band to the valence band, a 574-nm photon is emitted. What is the size of the band gap? Get solution

10q. Describe the effect of high temperatures on a semiconductor diode. Get solution

11. Find the ratio of forward-bias to reverse-bias currents when the same voltage 1.5 V is applied in both forward and reverse. Assume room temperature 293 K. Get solution

12. Suppose you want the ratio of forward to reverse bias current in a diode to be -106 at room temperature 293 K when the same voltage is applied in both forward and reverse. What voltage is required? Get solution

12q. Why is it appropriate to think of a photovoltaic cell as an LED operating in reverse? Get solution

13. Find the fraction of the standard solar flux reaching the Earth (about 1000 W/m2) available to a solar collector lying flat on the Earth’s surface at each of the following places at noon on the winter solstice, spring equinox, and summer solstice: (a) Miami, latitude 26° N; (b) Regina, Saskatchewan, latitude 50° N; and (c) St. Petersburg, Russia, latitude 60° N. Get solution

13q. What are some possible applications of carbon nanotubes? Get solution

14. Assuming that the average daily solar constant at a particular place is 200 W/m2, how large an array of 30% efficient solar cells is required to equal the power output of a typical power plant, about 109 W? Get solution

14q. Why are quantum effects important in electronic circuits that are nanometers in size but not in those circuits microns in size? Get solution

15. For the diode described in Example 11.4, find the forward bias current with V = 250 mV at (a) T = 250 K, (b) T = 300 K, and (c) T = 500 K. ... Get solution

15q. Explain why the size of the band gap in a quantum dot decreases as the size of the dot increases. Get solution

16. A single-walled carbon nanotube has 2.3 × 1019 carbon atoms per m2 along its surface. The nanotube diameter is 1.4 nm. (a) Find the mass density of the nanotube in kg/m3. (b) Compare your answer to (a) with the density of steel, about 7800 kg/m3. Get solution

17. Using the data in the preceding problem, find the density of a nanotube peapod that also contains one buckyball (C60) for each nm of its length. Get solution

18g. The Fermi-Dirac factor is expressed in Equation (9.34) as ... In a semiconductor or insulator, with an energy gap Eg between the valence and conduction bands, we can take EF to be halfway between the bands, so that EF = Eg/2. (a) Show that for a typical semiconductor or insulator at room temperature the Fermi-Dirac factor is approximately equal to exp (-Eg/2kT). (b) Use the result in (a) to compute the Fermi-Dirac factor for a typical insulator, with Eg = 8.0 eV at T = 300 K. (c) Repeat for a semiconductor, silicon, with Eg = 1.11 eV at T = 293 K. (d) Your result in (c) is still small, but sufficiently large to explain why there will be conduction. Explain. Get solution

19g. Consider what happens when silicon is doped with arsenic. Suppose that the extra, weakly bound electron from arsenic moves in the first Bohr orbit in a silicon atom. The first Bohr orbit has a radius of ... Because of the effects of screening, it is necessary to replace ε0 with the electric permittivity ..., where ...is the dielectric constant (... = 11.7 for silicon). (a) Compute the effective Bohr radius for this electron. (b) Compare your result with the lattice spacing in silicon, about 0.235 nm, and comment on the result. Get solution

20g. Follow the same procedure as in Problem 19 to find the binding energy E0 for the first Bohr orbit of the extra electron in silicon. Comment on the result. Consider what happens when silicon is doped with arsenic. Suppose that the extra, weakly bound electron from arsenic moves in the first Bohr orbit in a silicon atom. The first Bohr orbit has a radius of ... Because of the effects of screening, it is necessary to replace ε0 with the electric permittivity ... = 11.7 for silicon). (a) Compute the effective Bohr radius for this electron. (b) Compare your result with the lattice spacing in silicon, about 0.235 nm, and comment on the result. Get solution

21g. (a) Use the thermocouple data in Table 11.3 and a computer to perform a linear least-squares fi t of voltage versus temperature for the range 0°C to 50°C and for the range 50°C to 100°C. Over which range is the fi t better? (b) Repeat your analysis of (a), this time using a second-order least-squares fi t (voltage will be a function of T and T 2). Compare your results with those you obtained in (a). ... Get solution

22g. A certain diode has a reverse bias current of 1.05 A. Now this diode is connected in forward bias in the circuit shown below, in series with a resistor and with a constant voltage source of 6.0 volts. (a) Find the value of resistance R such that a current of 140 mA will flow at room temperature (293 K). (b) Under the condition described in (a), find the voltage drop across the resistor. ... Get solution

23g. Assume a temperature of 293 K and find the value V of the bias voltage in Equation (11.9) where (a) I = 7I0 and (b). ... ... Get solution

24g. A light-emitting diode made of the semiconductor GaAsP gives off red light (λ = 650 nm). Determine the energy gap for this semiconductor. Get solution

25g. How large an energy gap is required for a GaN laser used in a Blu-ray DVD player? Get solution

26g. (a) Find the length of each side of a square computer chip with 580 million transistors, if each transistor occupies a square of side 45 nm. (b) Find the number of transistors on a chip of the same size you found in (a) if the transistor size can be reduced to 32 nm on a side. Get solution

27g. Early research on semiconductor materials focused on silicon and germanium, which have band gaps of 1.11 eV and 0.67 eV, respectively. (a) Use the result of Problem 18a to compute the Fermi-Dirac factor for silicon and germanium at T = 0°C and T = 75°C. (This represents a fair temperature range for semiconductors in use.) On the basis of your computed value for FFD for germanium at the higher temperature, explain why silicon is preferred in most applications. Get solution

28g. Suppose the average solar flux reaching the United States is 200 W/m2. This average is taken over a whole year, and takes into account seasonal effects and weather (clouds), and assumes fixed solar cells. (a) Find the total energy produced in one year by a 1-m2 cell producing energy with an efficiency of 15%. (b) How much area would have to be covered with these solar cells to supply the United States with all its electricity, if the yearly electrical energy consumption in the United States is about 4.0 × 1012 kW . h? (c) Real solar arrays require about 2.5 times as much area as you found in (b) in order to keep one array of cells from shading another. What fraction of the land area of the United States (about 9 × 106 km2) would have to be covered with solar cells to meet the nation’s energy requirements? Get solution

29g. In this problem you will examine the temperature dependence of the forward/reverse bias modes of a diode. (a) For a pn-junction diode compute the ratio of forward bias current to reverse bias current with an applied voltage of 1.50 volts at each of the following temperatures: 77 K, 273 K, 350 K, 600 K. (b) Comment on the results with respect to possible applications. Get solution

30g. Pure silicon is used as a photon detector. An incoming photon can strike the surface and excite electrons from the valence band to the conduction band, where they can be counted. (a) Compute the number of electrons you would expect to count if a silicon detector is struck with a 1.04-MeV gamma ray produced in the decay of a 136Cs nucleus. (b) Explain why the counting of electrons should be more precise if the detector is cooled well below room temperature. Get solution

31g. Suppose you apply the same voltage, 250 mV, to a pn-junction diode, first in forward bias and then in reverse bias, at room temperature (293 K). What is the ratio of currents in forward and reverse bias? Get solution

32g. A carbon nanotube has a diameter of 1.6 nm. Young’s modulus for the nanotube is 1050 GPa. With one end of the tube fixed, how large a force must be applied to the other end to increase the tube’s length by 1%? Get solution

33g. A DVD has an effective inner radius of 2.3 cm and outer radius of 5.8 cm. The disk’s capacity is 4.7 gigabytes, where 1 byte = 8 bits. Find the number of bits stored per square meter, and compare with the density on a magnetic device, 1 = 1013 bits/m2. Get solution

34g. A Blu-ray DVD has the same dimensions as the DVD described in the preceding problem, but it can store 25 gigabytes of information. If the width of each track of the Blu-ray DVD is 320 nm, what is the average bit length? Get solution

35g. The Nevada Solar One solar thermal power plant covers a land area of 140 hectares and has an estimated peak output of 64 MW. If the peak solar flux reaching the surface of that part of Nevada is about 630 W/m2, what is the net efficiency of the power plant? Compare your answer with some of the higher efficiencies of semiconductor solar cells described in the text. Get solution

36g. A quantum dot is composed of CdS, with a density of 4820 kg/m3. (a) Find the number of atoms in a spherical quantum dot of radius 2.50 nm. (b) Model the energy levels of a quantum dot as a one-dimensional infinite potential well of 2.0-nm width. What is the lowest energy level in this well? (c) What is the “band gap” between the n = 5 and n = 6 levels? Get solution


Chapter #16 Solutions - Modern Physics for Scientists and Engineers - Andrew Rex, Stephen T. Thornton - 4th Edition

1. Derive the conversion from parsecs to lightyears given the information in Example 16.1. ... Get solution 1q. Explain why Hubbl...