Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 38, Problem 2P
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The energy gaps Eg for the semiconductors silicon and germanium are, respectively, 1.12 and 0.67 eV. Which of the following statements, if any, are true? (a) Both substances have the same number density of charge carriers at room temperature. (b) At room temperature, germanium has a greater number density of charge carriers than silicon. (c) Both substances have a greater number density of conduction electrons than holes. (d) For each substance, the number density of electrons equals that of holes.
(e) Intrinsic silicon has effective densities of states in the conduction band and the
valence band of 3.2 × 10¹⁹ cm−³ and 1.8 × 10¹⁹ cm-³, respectively. If the band gap
is 1.12 eV, what is the concentration of intrinsic charge carriers in silicon at 300 K?
A.
9.46 x 10⁹ m-³
9.46 x 10⁹ cm-³
0 m-3
2.40 x 1019 cm-3
B.
C.
D.
Iron has an atomic number Z-26, relative atomic mass of A=52 and a mass density of 7800 kg m-3, Estimate the number of conduction band electrons per m 3 in
iron, assuming that each atom contributes two electrons to the conduction band. Recall that m e-9.1*10-31 kg, m p=1.67*10-27 kg, e=1.6*10-19 C, N A-6.022*10 23
mol-1
O 17.202 1028
17.482-1028
17.964*1028
O
O 36.158 1028
O 58.725 1028
Chapter 38 Solutions
Physics for Scientists and Engineers
Ch. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10P
Ch. 38 - Prob. 11PCh. 38 - Prob. 12PCh. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - Prob. 19PCh. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Prob. 27PCh. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - Prob. 30PCh. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Prob. 33PCh. 38 - Prob. 34PCh. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53PCh. 38 - Prob. 54PCh. 38 - Prob. 55PCh. 38 - Prob. 56PCh. 38 - Prob. 57PCh. 38 - Prob. 58PCh. 38 - Prob. 59PCh. 38 - Prob. 60PCh. 38 - Prob. 61PCh. 38 - Prob. 62PCh. 38 - Prob. 63PCh. 38 - Prob. 64PCh. 38 - Prob. 65PCh. 38 - Prob. 66PCh. 38 - Prob. 67PCh. 38 - Prob. 68PCh. 38 - Prob. 69PCh. 38 - Prob. 70PCh. 38 - Prob. 71PCh. 38 - Prob. 72PCh. 38 - Prob. 73PCh. 38 - Prob. 74PCh. 38 - Prob. 75PCh. 38 - Prob. 76P
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- In an unbiased p-n junction, holes diffuse from the p-region to n-region because(a) free electrons in the n-region attract them.(b) they move across the junction by the potential difference.(c) hole concentration in p-region is more as compared to n-region.(d) All the abovearrow_forwardIn a pure ("intrinsic") semiconductor material is heated, the thermal energy liberates some valence-band electrons into the conduction band. The vacancies left behind in the valence band are called holes: If an electrical voltage is applied across the heated semiconducting substance, with positive on the left and negative on the right, what will this do to the energy bands, and how will this affect both the electrons and the holes?arrow_forwardPure silicon at room temperature has an electron number density in the conduction band of about 5 * 105 m-3 and an equal density of holes in the valence band. Suppose that one of every 107 silicon atoms is replaced by a phosphorus atom. (a) Which type will the doped semiconductor be, n or p? (b) What charge carrier number density will the phosphorus add? (c) What is the ratio of the charge carrier number density (electrons in the conduction band and holes in the valence band) in the doped silicon to that in pure silicon?arrow_forward
- 5. (a)How you conclude that the all superconductors are diamagnetic but all diamagnetic materials are not superconductors. (b)Determine the transition temperature and critical field at 4.2 K for a given specimen of a superconductor if the critical fields are 1.41 x 105 and 4.205 x 105 amp/m at 14.1 Kand 12.9 K, respectively.arrow_forwardIn a certain semiconductor, the valence band can be approximated by the function E(k) = Eo ak² and the conduction band can be described by the function E(k)= E₁ + 3k². Here E(k) is the electron energy and k is the wavevector. Plot E(k) for the two bands. What is the bandgap of this semiconductor? Is this a direct or indirect bandgap semiconductor?arrow_forwardA typical television remote control uses signal from an infrared LED. If the infrared light has a wavelength of 900 nm,(a). What is the energy band gap of the semiconductor in the LED? (b). What is the working principle of a semiconductor LED?arrow_forward
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