/*! 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} Q11P Nuclear radii may be measured by... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Nuclear radii may be measured by scattering high energy (high speed) electrons from nuclei. (a) What is the de-Broglie wavelength for 200MeV electrons? (b) Are these electrons suitable probes for this purpose?

Short Answer

Expert verified
  1. The de-Broglie wavelength for electrons is 6.2 fm.
  2. Yes, these electrons are suitable for the purpose of lower energy probes.

Step by step solution

01

The given data:

The energy of the given electrons, K = 200 MeV

02

Understanding the concept of the relativistic case:

You can get the detail of the nucleus if only the de-Broglie wavelength is smaller than that compared to the size or the radius of the nucleus. Since the kinetic energy K of the electron is much greater than its rest energy, a relativistic formulation must be used. Thus, using the relativistic approach to the concept of kinetic energy and momentum relation, we can get the required relativistic energy value that will further help in getting the de-Broglie wavelength.

Formulae:

The de-Broglie wavelength of a body,

λ=hp ….. (i)

Where, p is the momentum and h is Planck’s constant.

The relativistic relation between the energy and momentum,

pc=K2+m2c4 ….. (ii)

Here, m is the mass and c is the speed of light.

The kinetic energy of a body in motion with speed of light,

K=12mc2 ….. (iii)

03

(a) Calculation of the de-Broglie wavelength:

With the given data, equation (iii) and mc2=0.511MeVin equation (ii), you can get the value as follows:

pc=K2+2Kmc2=200MeV2+2200MeV0.511MeV=40000MeV2+204.4MeV2=200.5MeV

Thus, using the above value in equation (i) after multiplying in both the numerator and denominator, we can get the de-Broglie wavelength value as follows:

λ=hcpc=1240eV.nm200.5×106eV=6.18×10-6nm≈6.2fm

Hence, the value of wavelength is 6.2 fm.

04

(b) Calculation to know about the suitable electrons for the probe:

The diameter of a copper nucleus, for example, is about 8.6fm , just a little larger than the de Broglie wavelength of an 200 MeV electron. To resolve detail, the wavelength should be smaller than the target, ideally a tenth of the diameter or less. The 200 MeV electrons are perhaps at the lower limit in energy for useful probes.

The more energetic the incident particle; the finer are the details of the target that can be probed.

Hence, these electrons are suitable for the purpose of lower energy probes.

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

Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a neutron and a proton meet (assume both to be almost stationary), they combine and form a deuteron, emitting a gamma ray whose energy is 2.2233MeV. The masses of the proton and the deuteron are1.007276467uand 2.103553212u, respectively. Find the mass of the neutron from these data.

Consider an initially pure3.4g sample of67Ga, an isotope that has a half-life of 78h. (a) What is its initial decay rate? (b) What is its decay rate 48hlater?

A 5.00 gcharcoal sample from an ancient fire pit has an C14activity of 63.0disintegrations/min. A living tree has anactivity ofrole="math" localid="1661591390811" 15.3disintegrations/minper 1.00 g. The half-life of14Cis 5730 y. How old is the charcoal sample?

Under certain rare circumstances, a nucleus can decay by emitting a particle more massive than an alpha particle. Consider the decays Ra223→Pb209+C12andRa223→Ra219+He4

Calculate the Qvalue for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is 30.0 MeV. What is the barrier height for14Cemission ? (Be careful about the nuclear radii.) The needed atomic masses are

Ra223223.01850uC1414.00324uRa20920.98107uC44.00260uRa219219.00948u

What is the binding energy per nucleon of the americium isotope Am95244? Here are some atomic masses and the neutron mass.

Am95244244.064279uH11.007825un1.008665u

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.