Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Question
Chapter 38, Problem 28P
(a)
To determine
The Fermi energy for
(b)
To determine
The Fermi energy for
(c)
To determine
The Fermi energy for
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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 a solid,consider the energy level lying 0.7eV below fermi level. what is the probability of this level not being occupied by an electron at the room temperature?arrow_forwardWhy does the horizontal Line in the graph in Figure 9.12 suddenly stop at the Fermi energy? Figure 9.12 (a) Density of state for a free electron gas; (b) probability that a state is occupied at T = 0 K; (c) density if occupied states at T = 0 k.arrow_forwardTo 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_forward
- At what temperature, in terms of Tc, is the critical field of a superconductor one-half its value at T = 0 K?arrow_forwardFind the equilibrium electron and hole concentration and the location of the Fermi level for a silicon sample at27∘Cdoped uniformly with5×1015 cm−3phosphorus atoms and4×1015 cm−3boron atoms.arrow_forwardIn a solid, consider the energy level lying 0.4eV below Fermi level.What is Probability of this level not being occupied by an electron at the room temperature?arrow_forward
- Suppose a pure Si crystal has 5 × 1028 atoms m-3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that ni =1.5 × 1016 m-3.arrow_forwardFor silicon at T = 500 K with donor density N_D = 5* 10^{13} cm^ and acceptor density N_A = 0 calculate the equilibrium hole concentration in cm^{-3}. In this problem, you can assume the bandgap energy and effective masses are independent of temperature and use the room temperature values for them. Values within 5% error will be considered correct.arrow_forwardConsider a n-type Si crystal at room temperature (300K) doped with 6 x1016 cm-3 arsenic impurity atoms and with certain number of shallowholes. Find out the equilibrium electron concentration, hole concentrationand Fermi level EF with respect to Ei, and the conduction band edge EC.For Si at 300K, the value of ni is 1.45 x 1010 cm-3 and k = 1.38 x 10-23 J/K,1eV = 1.60 x 10-19J. The band gap energy, Eg, of Si is 1.2eV.Solution:n @ Nd = 6 x 1016 cm-3.In equilibrium condition, hole concentration = 3.5 x 103 cm-3.EF – EI = 0.396eVEC – EF = 0.164eV.arrow_forward
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