When would a density of states be needed: in a sum over states? in a sum over energies? in an integral over energies? in an integral over states?

Short Answer

Expert verified

Thus, among the choices, density of states is most useful when we consider integral over energies.

Step by step solution

01

Step 1:Definition of density

The amount per unit of length, area, or volume as. (A): a substance's mass per unit volume. (B): the distribution of an amount (such as mass, electricity, or energy) per unit, typically of space.

02

integral over energy

Density of states refer to the number of states at a specified energy per infinitesimal energy range It is useful in the calculation of the average energy in the classical limit where the quantum levels are non-discrete and hence, tinier than individual particle energies.

Thus, among the choices, density of states is most useful when we consider integral over energies.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Fermi velocityVfis defined byEF=12msF2, whereEFis the Fermi energy. The Fermi energy for conduction electron in sodium is 3 IV. (a) Calculate the Fermis velocity (b) What would be the wavelength of an electron with this velocity? (c) If each sodium atom contributes one conduction electron to the electron gas and sodium atom are spaced roughly0.37nmapart. If it is necessary, by the criteria of equation (9-43), to treat the conduction electron gas as quantum gas?

This problem investigates what fraction of the available charge must be transferred from one conductor to another to produce a typical contact potential. (a) As a rough approximation treat the conductors as10cmx 10 cm square plates2cm apart-a parallel-plate capacitors so thatq=CV , where C=σ0(0.01m2/0.02m). How much charge must be transferred from one plate to the other to produce a potential difference of 2V?(b) Approximately what fraction would this be of the total number of conduction electrons in a 100gpiece of copper. which has one conduction electron per atom?

The exact probabilities of equation (9-9) rest on the claim that the number of ways of addingNdistinct non-negative integer to give a total ofM is (M+N-1)![M!(N-1)!] . One way to prove it involves the following trick. It represents two ways that Ndistinct integers can add toM-9 and5, respectively. In this special case.

The X's represent the total of the integers, M-each row has 5. The 1'srepresent "dividers" between the distinct integers of which there will of course be N-Ieach row has8 . The first row says thatn1 is3 (three X'sbefore the divider between it andn2 ), n2is0 (noX's between its left divider withn1 and its right divider withn3 ),n3 ) is1 . n4throughn6 are0 , n7is1 , and n8and n9are0 . The second row says that n2is 2. n6is 1, n9is2 , and all othern are0 . Further rows could account for all possible ways that the integers can add toM . Argue that properly applied, the binomial coefficient (discussed in AppendixJ ) can be invoked to give the correct total number of ways for anyN andM .

Exercise 52 gives the Boltzmann distribution for the special case of simple harmonic oscillators, expressed in terms of the constant ε=Nhω0/(2s+1). Exercise 53 gives the Bose-Einstein and Fermi-Dirac distributions in that case. Consider a temperature low enough that we might expect multiple particles to crowd into lower energy states:kBT=15ε. How many oscillators would be expected in a state of the lowest energy,E=0? Consider all three-classically distinguishable. boson, and fermion oscillators - and comment on the differences.

The fact that a laser's resonant cavity so effectively sharpens the wavelength can lead to the output of several closely spaced laser wavelengths, called longitudinal modes. Here we see how. Suppose the spontaneous emission serving as the seed for stimulated emission is of wavelength , but somewhat fuzzy, with a line width of roughly0.001nm either side of the central value. The resonant cavity is exactly60cm long. (a) How many wavelengths fit the standing-wave condition'? (b) If only u single wavelength were desired, would change the length of the cavity help? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free