As we see in Figures 10.23, in a one dimensional crystal of finite wells, top of the band states closely resemble infinite well states. In fact, the famous particle in a box energy formula gives a fair value for the energies of the band to which they belong. (a) If for nin that formula you use the number of anitnodes in the whole function, what would you use for the box length L? (b) If, instead, the n in the formula were taken to refer to band n, could you still use the formula? If so, what would you use for L? (c) Explain why the energies in a band do or do not depend on the size of the crystal as a whole.

Short Answer

Expert verified
  1. The length L will take the length of the whole crystal.
  2. The use of L is the length of the single atom.
  3. The reason is explained

Step by step solution

01

Determine the formulas

Consider the formula for the relation between the energy of the particle in the box as follows:

E=n2h28mL2

Here, E is the energy, n is the number of antinodes, h is the planck’s constant and m is the mass.

02

Determine the answer for part (a)

Consider the crystal is considered as the box and the length L is taken as the length of the whole crystal. As the length L covers the all the n antinodes.

03

Determine the answer for part (b)

Consider the equation that determines the energy as:

E=(nN)2πh22md2

Resolve the equation as:

E=(n)2πh22mWN2E=(n)2πh22ma2

Hence, for n to be the band the use of L is the length of the single atom as the n corresponds to the number of antinodes in the single atom.

04

Determine the answer for part (c)

Consider that the band energy is clustered around the corresponding single atom energy and it do not depend on the number of atoms in the crystal this is why the band energy is independent of the size of the crystal.

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

A function f(α)is nonzero only in the region of width 2δcentered atα=0

f(α)={Cαδ0αδ

where C is a constant.

(a) Find and plot versus βthe Fourier transform A(β)of this function.

(b) The function ρα) might represent a pulse occupying either finite distance (localid="1659781367200" α=position) or finite time (α=time). Comment on the wave number if α=is position and on the frequency spectrum if αis time. Specifically address the dependence of the width of the spectrum on δ.

Exercise 81 obtained formulas for hydrogen like atoms in which the nucleus is not assumed infinite, as in the chapter, but is of mass,m1 whilem2is the mass of the orbiting negative charge. (a) What percentage error is introduced in the hydrogen ground-state energy by assuming that the proton is of infinite mass? (b) Deuterium is a form of hydrogen in which a neutron joins the proton in the nucleus, making the nucleus twice as massive. Taking nuclear mass into account, by what percent do the ground-state energies of hydrogen and deuterium differ?

Determine the Fourier transform A(k)of the oscillatory functionf(x) ) and interpret the result. (The identity cos(k0x)=12(e+ik0+eik0)may be useful.)

Here we investigate the link between nand l, reflected in equation (7-33). (a) Show that if a classical point charge were held in a circular orbit about a fixed point charge by the Coulomb force, its kinetic energy would be given by KE=e2/8πε0r (b) According to equation (7-30), the rotational kinetic energy in hydrogen is h2l(l+1)/2mr2. Of course, ris not well defined for a “cloud”, but by usingr=n2a0argue that the condition that l not exceed n is reasonable.

Obtain a semi empirical binding energy per nucleon formula. Using this as a guide, explain why the Coulomb force, which is only about1100as strong as the inter nucleon attraction for two protons "in contact" (cf. Table 11.2), would eventually have to become a dominant factor in large nuclei. Assume that Z,N and A increase in rough proportion to one another.

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