/*! 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 86 Due to annihilation of electron-... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 86

Due to annihilation of electron-positron of same kinetic energy \(0.95 \mathrm{MeV}\), a photon is produced which can also be produced by a photo- electron of energy \(E\), the possible value(s) of \(E\) is/are (mass of electron \(=9.1 \times 10^{-31} \mathrm{~kg}, e=1.6 \times 10^{-19}\) Coulomb) (A) \(1.02 \mathrm{MeV}\) (B) \(2.42 \mathrm{MeV}\) (C) \(4.03 \mathrm{MeV}\) (D) \(2.93 \mathrm{MeV}\)

Short Answer

Expert verified
The possible value of E for the photon produced by a photo-electron is \(2.93 \mathrm{MeV}\).

Step by step solution

01

Electron-Positron Annihilation Energy/Momentum Conservation

When an electron and a positron annihilate, they release energy in the form of a photon. The energy-momentum conservation implies that the total energy and momentum before and after the annihilation must be the same. In this case, the kinetic energy of the electron and positron is given as 0.95 MeV. We can convert this energy to its mass-equivalent using Einstein's mass-energy relation, \(E = mc^2\), where m is mass, c is the speed of light, and E is the energy. To obtain the mass equivalent of the kinetic energy, we have: \[m_{electron} = E_{electron}/c^2 = 0.95 \times 10^6 \mathrm{eV} / (3 \times 10^8 \mathrm{m/s})^2\] \[m_{electron} = 1.698 \times 10^{-30} \mathrm{kg}\]
02

Mass of Annihilated Particle Pair

Now that we have the mass of the electron due to its kinetic energy, we can find the total mass of the electron-positron pair which can be converted into energy during the annihilation. Considering that the electron and the positron have the same mass and same kinetic energy, the total mass of the particle pair can be obtained as follows: \[Total \ Mass = m_{electron} + m_{positron} = 2 \times m_{electron} = 2 \times 1.698 \times 10^{-30} \mathrm{kg}\] \[Total \ Mass = 3.396 \times 10^{-30} \mathrm{kg}\]
03

Photon Energy

Now that we have the total mass of the annihilated particle pair, we can find the energy of the photon produced in the process using the mass-energy relation \(E = mc^2\): \[E_{photon} = Total \ Mass \times c^2 = 3.396 \times 10^{-30} \mathrm{kg} \times (3 \times 10^8 \mathrm{m/s})^2\] \[E_{photon} = 3.05 \mathrm{MeV}\]
04

Find the Possible Value(s) of E from the Given Options

Now that we know the energy of the photon produced in the electron-positron annihilation, we can compare it to the energies given in the options and find the possible value(s) of E: (A) \(1.02 \mathrm{MeV}\) - Not equal to our calculated E. (B) \(2.42 \mathrm{MeV}\) - Not equal to our calculated E. (C) \(4.03 \mathrm{MeV}\) - Higher than our calculated E. (D) \(2.93 \mathrm{MeV}\) - Closest to our calculated E. Based on our calculation, the possible value of E for the photon produced by a photo-electron is: (D) \(2.93 \mathrm{MeV}\)

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

The minimum \(K E\) of proton incident on \({ }_{6}^{13} \mathrm{C}\) nuclei at rest that will produce the reaction \({ }^{13} \mathrm{C}(p, n)^{13} \mathrm{~N}\) is (Mass of nitrogen \(=13.005738\) amu, Mass of neutron \(=1.008665\) amu, Mass of carbon \(=13.005738\) amu Mass of proton \(=1.007825 \mathrm{amu}\) ) (A) \(0.02 \mathrm{MeV}\) (B) \(0.10 \mathrm{MeV}\) (C) \(0.03 \mathrm{MeV}\) (D) \(0.05 \mathrm{MeV}\)

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) \(19.2 \mathrm{MeV}\)

Electromagnetic radiation whose electric component varies with time as \(E=C_{1}\left(C_{2}+C_{3} \cos \omega t\right) \cos \omega_{0} t\), here \(C_{1}, C_{2}\) and \(C_{3}\) are constants, is incident on lithium and liberates photoelectrons. Let the kinetic energy of most energetic electrons be \(0.592 \times 10^{-19} \mathrm{~J}\). Given that \(\omega_{0}=3.6 \times 10^{15} \mathrm{rad} / \mathrm{s}\) and \(\omega=6 \times 10^{14}\) \(\mathrm{rad} / \mathrm{sec} .\) The work function of lithium is (take planks constant \(\left.h=6.6 \times 10^{-34} \mathrm{MKS}\right)\). (A) \(1.2 \mathrm{eV}\) (B) \(1.5 \mathrm{eV}\) (C) \(2.1 \mathrm{eV}\) (D) \(2.39 \mathrm{eV}\)

The half-life of a radioactive substance is \(20 \mathrm{~min}\). The approximate time interval \(\left(t_{2}-t_{1}\right)\) between the time \(t_{2}\) when \(\frac{2}{3}\) of it has decayed and time \(t_{1}\) when \(\frac{1}{3}\) of it had decayed is (A) \(7 \mathrm{~min}\) (B) \(14 \mathrm{~min}\) (C) \(20 \mathrm{~min}\) (D) \(28 \mathrm{~min}\)

The transition from the state \(n=4\) to \(n=3\) in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from: (A) \(3 \rightarrow 2\) (B) \(4 \rightarrow 2\) (C) \(5 \rightarrow 4\) (D) \(2 \rightarrow 1\)
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Famous Physicists

Read Explanation

Space Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electric Charge Field and Potential

Read Explanation

Geometrical and Physical Optics

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.