Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 2542

A gamma ray Photon creates an electron-Positron Pair. If the rest mass energy of an electron is \(0.5 \mathrm{MeV}\). and the total kinetic energy $0.7 \mathrm{MeV}$, then the energy of the gamma ray Photon must be (A) \(3.9 \mathrm{MeV}\) (B) \(1.78 \mathrm{MeV}\) (C) \(0.78 \mathrm{MeV}\) (D) \(0.16 \mathrm{MeV}\)

Short Answer

Expert verified
The energy of the gamma-ray photon is approximately \(1.7\,\mathrm{MeV}\), which is closest to option (B) \(1.78\,\mathrm{MeV}\).

Step by step solution

01

Determine the Rest Mass Energy of the Electron-Positron Pair

We are given that the rest mass energy of an electron is \(0.5\,\mathrm{MeV}\). Since an electron and a positron have the same rest mass energy, the total rest mass energy of the pair is: Total rest mass energy = Rest mass energy of electron + Rest mass energy of positron = \(0.5\,\mathrm{MeV} + 0.5\,\mathrm{MeV} = 1.0\,\mathrm{MeV}\)
02

Determine the Total Energy of the Electron-Positron Pair

Now, we need to find the total energy of the electron-positron pair including their kinetic energy. We are given that the total kinetic energy of the pair is \(0.7\,\mathrm{MeV}\). Therefore, the total energy of the electron-positron pair is: Total energy = Total rest mass energy + Total kinetic energy = \(1.0\,\mathrm{MeV} + 0.7\,\mathrm{MeV} = 1.7\,\mathrm{MeV}\)
03

Apply the conservation of energy

According to the principle of conservation of energy, the energy of the gamma-ray photon must be equal to the total energy of the electron-positron pair. Therefore, the energy of the gamma-ray photon is: Energy of gamma-ray photon = Total energy of electron-positron pair = \(1.7\,\mathrm{MeV}\)
04

Identify the correct option

Now that we have calculated the energy of the gamma-ray photon, we can compare it with the given options: (A) \(3.9\,\mathrm{MeV}\) (B) \(1.78\,\mathrm{MeV}\) (C) \(0.78\,\mathrm{MeV}\) (D) \(0.16\,\mathrm{MeV}\) Since the energy of the gamma-ray photon is \(1.7\,\mathrm{MeV}\), it is closest to option (B), which is \(1.78\,\mathrm{MeV}\). Hence, the correct answer is: (B) \(1.78\,\mathrm{MeV}\)

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.

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

(i) statement-I :- Large angle scattering of alpha Particle led to discovery of atomic nucleus. statement-II :- Entire Positive charge of atom is concentrated in the central core. (A) statement -I and II are true. and statement II is correct explanation of statement-I (B) statement -I and II are true, but statement-II is not correct explanation of statement I (C) statement I is true, but statement II is false. (D) statement I is false but statement II is true (ii) statement-I \(1 \mathrm{amu}=931.48 \mathrm{MeV}\) statement-II It follows form \(E=m c^{2}\) (iii) statement -I:- half life time of tritium is \(12.5\) years statement-II:- The fraction of tritium that remains after 50 years is \(6.25 \%\) (iv) statement-I:- Nuclei of different atoms have same size statement-II:- \(\mathrm{R}=\operatorname{Ro}(\mathrm{A})^{1 / 3}\)

A Free neutron decays into a Proton, an electron and (A) \(\mathrm{v}\) (B) \(\underline{\mathrm{V}}\) (C) \(\beta\) (D) \(\alpha\)

two deuterons each of mass \(\mathrm{m}\) fuse to form helium resulting in release of energy \(\mathrm{E}\) the mass of helium formed is (A) \(\mathrm{m}+\left(\mathrm{E} / \mathrm{C}^{2}\right)\) (B) \(\left[\mathrm{E} /\left(\mathrm{mC}^{2}\right)\right]\) (C) \(2 \mathrm{~m}-\left(\mathrm{E} / \mathrm{C}^{2}\right)\) (D) \(2 \mathrm{~m}+\left(\mathrm{E} / \mathrm{C}^{2}\right)\)

The radio of minimum to maximum wave length in Balmer series is (A) \((1 / 4)\) (B) \((5 / 36)\) (C) \((3 / 4)\) (D) \((5 / 9)\)

In the following disintegration series ${ }_{92} \mathrm{U}^{238} \rightarrow^{\alpha} \mathrm{x} \rightarrow^{\beta-}{ } \mathrm{Z} \mathrm{y}^{\mathrm{A}}$ The value of \(Z\) and \(A\) respectively will be (A) 90,234 (B) 92,236 (C) 88,234 (D) 91,234
See all solutions

Recommended explanations on English Textbooks

Discourse

Read Explanation

Semiotics

Read Explanation

Sociolinguistics

Read Explanation

5 Paragraph Essay

Read Explanation

Lexis and Semantics Summary

Read Explanation

Email

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