/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q86E Prove that fur any sine function... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Prove that fur any sine function sin(kx+ϕ)of wavelength shorter than 2a, where ais the atomic spacing. there is a sine function with a wavelength longer than 2a that has the same values at the points x = a , 2a , 3a . and so on. (Note: It is probably easier to work with wave number than with wavelength. We sick to show that for every wave number greater than there is an equivalent less than π/a.)

Short Answer

Expert verified

It is proved that there is a sine function with a wavelength less than 2a.

Step by step solution

01

Sum of angle and expression for θ:

The atomic spacing is a.

The expression of sum of angles is given by,

θ1+θ2=nπ

The expression for is given by,

θ=kx+ϕ

02

Determine sum of angles:

The expression of sum of angles is calculated as,

θ1+θ2=nπnk1x+ϕ+k2x+ϕ=nπnk1x+k2x+2ϕ=nπn..........(1)

For

πana+ϕ=nπ2ϕ=nπ2-nπ=-nπ2

03

solve further:

Substitute -nπ2for ϕin equation (1).

k1+k2x+2-nÏ€2=nÏ€k1+k2na=2²ÔÏ€k1+k2=2Ï€ak1=2Ï€a-k2

Now since,

k1>Ï€a,2Ï€a-k2>Ï€a,k2<Ï€a

Therefore, it is proved that there is sine function with wavelength less than 2a.

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 91Ó°ÊÓ!

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

Question: Consider an electron in the ground state of a hydrogen atom. (a) Calculate the expectation value of its potential energy. (b) What is the expectation value of its kinetic energy? (Hint: What is the expectation value of the total energy?)

A particle is subject to a potential energy that has an essentially infinitely high wall at the origin, like the infinite well, but for positive values of x is of the form U(x)= -b/ x, where b is a constant

(a) Sketch this potential energy.

(b) How much energy could a classical particle have and still be bound by such a potential energy?

(c) Add to your sketch a plot of E for a bound particle and indicate the outer classical tuning point (the inner being the origin).

(d) Assuming that a quantum-mechanical description is in order, sketch a plausible ground-state wave function, making sure that your function's second derivative is of the proper sign when U(x)is less than E and when it is greater.

Question: Is the potential energy of an electron in a hydrogen atom well defined? Is the kinetic energy well defined? Justify your answers. (You need not actually calculate uncertainties.)

(a) Find the wavelength of a proton whose kinetic energy is equal 10 its integral energy.

(b) ' The proton is usually regarded as being roughly of radius10-15m. Would this proton behave as a wave or as a particle?

Nuclei of the same mass number but different Zare known as isobars. Oxygen-15 and nitrogen- 15 are isobars.

(a) In which of the factors considered in nuclear binding (represented by terms in the semi empirical binding energy formula) do these two isobars differ?

(b) Which of the isobars should be more tightly bound?

(c) k your conclusion in part (b) supported by the decay mode information of Appendix 1? Explain.

(d) Calculate the binding energies of oxygen-15 and nitrogen-15. By how much do they differ?

(e) Repeat part (d) but use the semi empirical binding energy formula rather than the known atomic masses.

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.