University Physics with Modern Physics (14th Edition)
14th Edition
ISBN: 9780321973610
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
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Question
Chapter 42, Problem 42.51P
(a)
To determine
To plot: The graph of energy
(b)
To determine
The temperature
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Chapter 42 Solutions
University Physics with Modern Physics (14th Edition)
Ch. 42.1 - If electrons obeyed the exclusion principle but...Ch. 42.2 - Prob. 42.2TYUCh. 42.3 - Prob. 42.3TYUCh. 42.4 - One type of thermometer works by measuring the...Ch. 42.5 - Prob. 42.5TYUCh. 42.6 - Prob. 42.6TYUCh. 42.7 - Suppose a negative charge is placed on the gate of...Ch. 42 - Van der Waals bonds occur in many molecules, but...Ch. 42 - Prob. 42.2DQCh. 42 - The H2+ molecule consists of two hydrogen nuclei...
Ch. 42 - The moment of inertia for an axis through the...Ch. 42 - Prob. 42.5DQCh. 42 - Prob. 42.6DQCh. 42 - Prob. 42.7DQCh. 42 - The air you are breathing contains primarily...Ch. 42 - Prob. 42.9DQCh. 42 - Prob. 42.10DQCh. 42 - What factors determine whether a material is a...Ch. 42 - Prob. 42.12DQCh. 42 - Prob. 42.13DQCh. 42 - Prob. 42.14DQCh. 42 - Prob. 42.15DQCh. 42 - Prob. 42.16DQCh. 42 - Prob. 42.17DQCh. 42 - Prob. 42.18DQCh. 42 - Prob. 42.19DQCh. 42 - Prob. 42.20DQCh. 42 - Prob. 42.21DQCh. 42 - Prob. 42.22DQCh. 42 - Prob. 42.23DQCh. 42 - Prob. 42.24DQCh. 42 - If the energy of the H2 covalent bond is 4.48 eV,...Ch. 42 - An Ionic Bond, (a) Calculate the electric...Ch. 42 - Prob. 42.3ECh. 42 - Prob. 42.4ECh. 42 - Prob. 42.5ECh. 42 - Prob. 42.6ECh. 42 - Prob. 42.7ECh. 42 - Two atoms of cesium (Cs) can form a Cs2 molecule....Ch. 42 - Prob. 42.9ECh. 42 - Prob. 42.10ECh. 42 - A lithium atom has mass 1.17 1026 kg, and a...Ch. 42 - Prob. 42.12ECh. 42 - When a hypothetical diatomic molecule having atoms...Ch. 42 - The vibrational and rotational energies of the CO...Ch. 42 - Prob. 42.15ECh. 42 - Prob. 42.16ECh. 42 - Prob. 42.17ECh. 42 - Prob. 42.18ECh. 42 - Prob. 42.19ECh. 42 - Prob. 42.20ECh. 42 - Prob. 42.21ECh. 42 - Prob. 42.22ECh. 42 - Prob. 42.23ECh. 42 - Prob. 42.24ECh. 42 - Prob. 42.25ECh. 42 - Prob. 42.26ECh. 42 - Prob. 42.27ECh. 42 - Prob. 42.28ECh. 42 - Prob. 42.29ECh. 42 - Prob. 42.30ECh. 42 - Prob. 42.31ECh. 42 - Prob. 42.32ECh. 42 - Prob. 42.33PCh. 42 - Prob. 42.34PCh. 42 - Prob. 42.35PCh. 42 - The binding energy of a potassium chloride...Ch. 42 - (a) For the sodium chloride molecule (NaCl)...Ch. 42 - Prob. 42.38PCh. 42 - Prob. 42.39PCh. 42 - Prob. 42.40PCh. 42 - Prob. 42.41PCh. 42 - Prob. 42.42PCh. 42 - Prob. 42.43PCh. 42 - Prob. 42.44PCh. 42 - Prob. 42.45PCh. 42 - Prob. 42.46PCh. 42 - Prob. 42.47PCh. 42 - Prob. 42.48PCh. 42 - Prob. 42.49PCh. 42 - Prob. 42.50PCh. 42 - Prob. 42.51PCh. 42 - Prob. 42.52PCh. 42 - Prob. 42.53CPCh. 42 - Prob. 42.54CPCh. 42 - Prob. 42.55CPCh. 42 - Prob. 42.56PPCh. 42 - Prob. 42.57PPCh. 42 - Prob. 42.58PP
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- To obtain a more clearly defined picture of the FermiDirac distribution, consider a system of 20 FermiDirac particles sharing 94 units of energy. By drawing diagrams like Figure P10.11, show that there are nine different microstates. Using Equation 10.2, calculate and plot the average number of particles in each energy level from 0 to 14E. Locate the Fermi energy at 0 K on your plot from the fact that electrons at 0 K fill all the levels consecutively up to the Fermi energy. (At 0 K the system no longer has 94 units of energy, but has the minimum amount of 90E.) 1 Microstate8 others? One of the nine equally probable microstates for 20 FD particles with a total energy of 94E.arrow_forwardFor the free electrons in a solid, what is the value of the E/EF ratio knowing that the occupancy factor of the energy level E is equal to 0.05 at a temperature such that kT=EF/4? (NOTE: EF is the Fermi energy).arrow_forwardThe Fermi energy level for a particular material at T = 300 K is 5.50 eV. The electrons in this material follow the Fermi-Dirac distribution function. a) Find the probability of an energy level at 5.50 eV being occupied by an electron. b) Repeat part (a) if the temperature is increased to T = 600 K. (Assume that EF is a constant.). c) Calculate the energy level where probability of finding an electron at room temperature is 70%. d) Calculate the temperature at which there is a 7 percent probability that a state 0.4 eV below the Fermi level will be empty of an electron.arrow_forward
- Given the fermi energy and electron concentration 7.00 eV and 8.0×10²6 e¯/m³ respectively of a Copper of resistivity 1.7×108 2-m, calculate the mean free path. (a) 3780 nm (b) 5000 nm (c) 4100 nm (d) 7000 nmarrow_forwardThe 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.arrow_forwardFor a solid metal having a Fermi energy of 8.500 eV, what is the probability, at room temperature, that a state having an energy of 8.520 eV is occupied by an electron?arrow_forward
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