(a) Explanation Annihilation of proton and positron to given two gamma rays can be represented as shown below. e − 1 0 + β + 1 0 → 2 γ Mass of one electron is 0 .00054 u and that of positron is 0 .00054 u . The mass defect for the annihilation reaction can be calculated using the equation given below. Δ m = m R − m P (1) Where, Δ m is the mass defect. m R is the mass of reactants. m P is the mass of products. Substituting the required values in equation (1), the mass defect is calculated as shown below. Δ m = m R − m P = 0.00054 u + 0.00054 u − 0 = 0 .00108 u Mass defect calculated in atomic mass unit is converted into grams by use of conversion factor as shown below. 0.00108 u = 0.00108 u × 1.6605 × 10 − 27 kg 1 u = 0 .00179334 × 10 − 27 kg = 1.7933 × 10 − 30 kg Energy released during the annihilation process can be calculated as shown below (b) Interpretation Introduction Interpretation: Frequency of gamma rays has to be calculated using the Planck’s equation.

3rd Edition

ISBN: 9780534420123

Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher: Cengage Learning

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Question

Chapter 21, Problem 21.60QE

(a)

Interpretation Introduction

Interpretation:

From the annihilation of electron and positron, the energy evolved has to be calculated.

(b)

Interpretation Introduction

Interpretation:

Frequency of gamma rays has to be calculated using the Planck’s equation.

Expert Solution & Answer

Chapter 21 Solutions

Chemistry: Principles and Practice

Ch. 21 - Prob. 21.1QE Ch. 21 - Prob. 21.2QE Ch. 21 - Prob. 21.3QE Ch. 21 - Prob. 21.4QE Ch. 21 - Prob. 21.5QE Ch. 21 - Prob. 21.6QE Ch. 21 - Prob. 21.7QE Ch. 21 - Prob. 21.8QE Ch. 21 - Prob. 21.9QE Ch. 21 - Prob. 21.10QE

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Define the term half-life and illustrate it with an example.

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