/*! 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 Calculate the disintegration ene... [FREE SOLUTION] | 91影视

91影视

Calculate the disintegration energy Q for the fission of C52rinto two equal fragments. The masses you will need are

role="math" localid="1661753124790" C52r51.94051M26g25.98259u

Short Answer

Expert verified

The disintegrated energy is -23MeV.

Step by step solution

01

Given data

The mass of C52r,mcr=51.94051u

The mass ofM26g,mMg=25.98259u

02

Determine the formula to calculate the disintegrated energy.

The expression to calculate the disintegrated energy is given as follows.

Q=mc2Q=(mCr2mMg)c2 ...(i)

03

Calculate the value of disintegrated energy.

Consider the fission reaction as given follow.

C52rM26g+M26g

Calculate the disintegrated energy.

Substitute51.94051u for mCr,25.98259u for MMgand931.5MeV/u forc2 into equation (i).

Q=51.9405u-225.98259u931.5MeV/uQ=-0.02467931.5MeVQ=-23MeV

Hence the disintegrated energy is -23MeV.

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 91影视!

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 isotope 235Udecays by alpha emission with a half-life of 7108y. It also decays (rarely) by spontaneous fission, and if the alpha decay did not occur, its half-life due to spontaneous fission alone would be 31017y.

(a) At what rate do spontaneous fission decays occur in 1.0 g of235U ?

(b) How many 235U alpha-decay events are there for every spontaneous fission event?

Calculate and compare the energy released by (a) the fusion of1.0 kg of hydrogen deep within the Sun and (b) the fission of 1.0 kgofU235in a fission reactor.

In an atomic bomb, energy release is due to the uncontrolled fission of plutonium Pu239(or U235). The bomb鈥檚 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 2.61028MeVof 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?

The fission properties of the plutonium isotope Pu239are very similar to those of U235. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1.0 kg of purePu239 undergo fission?

Question:(a) A neutron of mass mnand kinetic energy K makes a head-on elastic collision with a stationary atom of mass . Show that the fractional kinetic energy loss of the neutron is given by KK=4mnm(m+mn)2.

Find role="math" localid="1661942719139" KKfor each of the following acting as the stationary atom:

(b) hydrogen,

(c) deuterium,

(d) carbon, and

(e) lead.

(f) If K=1.00MeV initially, how many such head-on collisions would it take to reduce the neutron鈥檚 kinetic energy to a thermal value (0.25 eV) if the stationary atoms it collides with are deuterium, a commonly used moderator? (In actual moderators, most collisions are not head-on.)

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.