Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 43: Q53P (page 1333)

Verify that, as stated in Module 43-1, neutrons in equilibrium with matter at room temperature, 300 K, have an average kinetic energy of about 0.04 eV.

Short Answer

Expert verified

It is proved that the average kinetic energy is 0.04 eV.

Step by step solution

01

Describe the expression for average kinetic energy

The expression for the average kinetic energy is given by,

Kavg=32KT …… (1)

Here, T is temperature in kelvin.

02

Find the average kinetic energy

Substitute all the known values in equation (1).

Kavg=328.62×10-5eV/K300K=0.04eV

Therefore, it is proved that the average kinetic energy is 0.04 eV.

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.

Start your free trial

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

Coal burns according to the reaction C+O2→CO2. The heat of combustion is 3.3×107j/kgof atomic carbon consumed. (a) Express this in terms of energy per carbon atom. (b) Express it in terms of energy per kilogram of the initial reactants, carbon and oxygen. (c) Suppose that the Sun (mass=2.0×1030kg) were made of carbon and oxygen in combustible proportions and that it continued to radiate energy at its present rate of 3.9×1026W. How long would the Sun last?

The nuclide238Np requires4.2 MeVfor fission. To remove a neutron from this nuclide requires an energy expenditure of 5.0 MeV. Isfissionable by thermal neutrons?

In an atomic bomb, energy release is due to the uncontrolled fission of plutonium Pu239(or U235). The bomb’s rating is the magnitude of the released energy, specified in terms of the mass of TNT required to produce the same energy release. One megaton of TNT releases 2.6×1028MeVof energy. (a) Calculate the rating, in tons of TNT, of an atomic bomb containing 95 kg of Pu239, of which 2.5 kg actually undergoes fission. (See Problem 4.) (b) Why is the other 92.5 kg of Pu239needed if it does not fission?

(a) Calculate the rate at which the Sun generates neutrinos. Assume that energy production is entirely by the proton-proton fusion cycle. (b) At what rate do solar neutrinos reach Earth?

In Fig. 43-10, the equation for n(K), the number density per unit energy for particles, n(K)=1.13nK1/2(kT)3/2e-KkTiswhere nis the total particle number density. At the center of the Sun, the temperature is1.50×107Kand the average proton energyKavgis 1.94 keV. Find the ratio of the proton number density at 5.00keVto the number density at the average proton energy.
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Electricity and Magnetism

Read Explanation

Rotational Dynamics

Read Explanation

Engineering Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Work Energy and Power

Read Explanation
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