Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
expand_more
expand_more
format_list_bulleted
Question
Chapter 38, Problem 60P
To determine
Prove that the Fermi factor is equal to
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Using the Fermi function, estimate the temperature at which there is a 1% probability
that an electron in a solid will have an energy of 0.5 eV above the Fermi energy.
Compute the Fermi speed for (a) Ca (EF = 4.69 eV) and (b) Be (EF = 14.3 eV).
Plot the Fermi function Vs. Energy at the temperature of 500 K, when EF = 2 eV
Chapter 38 Solutions
Physics for Scientists and Engineers, Vol. 1
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
Knowledge Booster
Similar questions
- 48 Show that P(E), the occupancy probability in Eq. 41-6, is sym- metrical about the value of the Fermi energy; that is, show that P(EF + AE) + P(EF - AE) = 1.arrow_forwardEstimate the fraction of electrons excite above the Fermi level at room temperature (T=300K) for Sodium and copper? Given the Fermi energy (EF) for sodium and copper is 3.1eV and 7eV respectively.arrow_forwardA state 63 meV above the Fermi level has a probability of occupancy of 0.090.What is the probability of occupancy for a state 63 meV below the Fermi level?arrow_forward
- 10-18. Consider a system in which the density of states of the electrons f(ɛ) is f(E) = constant = D ɛ>0 ɛ<0 Calculate the Fermi energy for this system; determine the condition for the system being highly degenerate; and then show that the heat capacity is proportional to T for the highly degenerate case.arrow_forwardAn atom’s nucleus is a collection of fermions— protons and neutrons. (a) In calculating the Fermi energy in a nucleus, the protons and neutrons must be considered separately. Why? (b) Find the Fermi energy of (i) the protons and (ii) the neutrons in a uranium nucleus, which has a radius of 7.4 x 10-15 m and contains 92 protons and 146 neutrons.arrow_forwardLet f(ɛ) be the Fermi-Dirac distribution function and u be the chemical potential. Obtain the expression for the derivative of f (ɛ)with respect to ɛat & = u . 1 (a) 27 1 (b) 6t 1 (c) 47 1 (d) - -arrow_forward
- The probability that a state at Ec+kT is occupied by an electron is equal to the probability that a state at Ey-kT is empty. Determine the position of the Fermi energy level as a function of Ec and Ey.arrow_forwardProve that mean energy of the electrons at absolute zero <E> = 3.Ef/5 where Ef is the Fermi energy. Show also that <v^2>/<v>^2 = 16/15arrow_forwardIn a Si semiconductor sample of 200 am length at 600 K the hole concentration as a' function of the sample length follows a quadratic relation of the form p (x) = 1 x1015x, at equilibrium the value of the electric field at 160 jum will be: O 1.935 V/cm O 3.250 V/cm O 5805 V/cm O 55.56 V/cm O 6.450 V/cmarrow_forward
- Consider a sample of GaAs at 300 K in which the Fermi level is 0.40 eV below the bottomof the conduction band. For the following questions, the Boltzmann approximation isvalid.a) What is the probability the energy state Ec is occupied by an electron?b) What is the probability a state E = 0.20 eV above the valence band is empty?c) What is the carrier concentration of electrons in this sample? Holes?arrow_forwardPlot the Fermi-Dirac function f(E) versus the energy ratio E/EF at room temperature T=300oK.(Assume EF independent of temperature.)If EF=5ev, determine the energy values at which f (E)=0.5.0.7.0.9 and 0.95?arrow_forwardStarting with the Fermi energy given in Table , estimate the number of conduction electrons per atom for aluminum, which has density 2.70 x 103 kg/m3 at T = 300 Karrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning