/*! 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} Problem 19 If the binding energy per nucleo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 19

If the binding energy per nucleon in \({ }_{3}^{7} \mathrm{Li}\) and \({ }_{2}^{4} \mathrm{He}\) nuclei are \(5.60 \mathrm{MeV}\) and \(7.06 \mathrm{MeV}\) respectively, then in the reaction \({ }_{1}^{1} \mathrm{H}+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}\) energy of proton must be (A) \(39.2 \mathrm{MeV}\) (B) \(28.24 \mathrm{MeV}\) (C) \(17.28 \mathrm{MeV}\) (D) \(1.46 \mathrm{MeV}\)

Short Answer

Expert verified
The energy of the proton (\({}_{1}^{1}\mathrm{H}\)) in the reaction provided is \(17.28 \mathrm{MeV}\). This corresponds to option (C) in the given choices.

Step by step solution

01

鈥� Calculate Initial Total Energy

First calculate initial total energy of proton and Lithium. For Lithium (\({ }_{3}^{7}\mathrm{Li}\)) nucleus, we know that the binding energy per nucleon is 5.60 MeV. Total energy is calculated by multiplying energy per nucleon with number of nucleons (7 nucleons in this case). The proton (\({}_{1}^{1}\mathrm{H}\)) has no binding energy since it's just one nucleon. So initial total energy is \(5.60 \mathrm{MeV}\times 7 + 0 = 39.2 \mathrm{MeV}\)
02

- Calculate Final Total Energy

Next, calculate the final total energy. In the product side, there are 2 Helium (\({}_{2}^{4} \mathrm{He}\)) nuclei each with 4 nucleons. The binding energy per nucleon is given as 7.06 MeV. Total final energy is \(2\times 4 \times 7.06\mathrm{MeV}=56.48\mathrm{MeV}\).
03

- Calculate the Energy of Proton

From conservation of energy, the total energy before the reaction is equal to total energy after the reaction. So if \(E_{p}\) represents the energy of the proton, then \(E_{p} + 39.2\mathrm{MeV} = 56.48\mathrm{MeV}\). Simplifying gives us \(E_{p} = 56.48\mathrm{MeV} - 39.2\mathrm{MeV} = 17.28\mathrm{MeV}\).

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

A radioactive sample at any instant has its disintegration rate 5000 disintegration per minute. After 5 minute, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is (A) \(0.4 \ln 2\) (B) \(0.2 \ln 2\) (C) \(0.1 \ln 2\) (D) \(0.8 \ln 2\)

In Bohr's model, the atomic radius of the first orbit is \(r_{0}\), then the radius of the third orbit is (A) \(\frac{r_{0}}{9}\) (B) \(r_{0}\) (C) \(9 r_{0}\) (D) \(3 r_{0}\)

In an electromagnetic wave, the electric and magnetizing fields are \(100 \mathrm{Vm}^{-1}\) and \(0.265 \mathrm{Am}^{-1}\). The maximum energy flow is (A) \(26.5 \mathrm{~W} / \mathrm{m}^{2}\) (B) \(36.5 \mathrm{~W} / \mathrm{m}^{2}\) (C) \(46.7 \mathrm{~W} / \mathrm{m}^{2}\) (D) \(765 \mathrm{~W} / \mathrm{m}^{2}\)

The binding energy per nucleon of deuteron \(\left({ }_{1}^{2} \mathrm{H}\right)\) and helium nucleus \(\left({ }_{2}^{4} \mathrm{He}\right)\) is \(1.1 \mathrm{MeV}\) and \(7 \mathrm{MeV}\) respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is (A) \(23.6 \mathrm{MeV}\) (B) \(26.9 \mathrm{MeV}\) (C) \(13.9 \mathrm{MeV}\) (D) \(19.2 \mathrm{MeV}\)

As an electron makes a transition from an excited state to the ground state of hydrogen - like atom/ion \([2015]\) (A) \(K E\), potential energy and total energy decrease. (B) \(K E\) decreases, potential energy increases but total energy remains same. (C) \(K E\) and total energy decrease but potential energy increases. (D) Its kinetic energy increases but potential energy and total energy decrease.
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Force

Read Explanation

Classical Mechanics

Read Explanation

Oscillations

Read Explanation

Physical Quantities And Units

Read Explanation

Rotational Dynamics

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.