Explanation Given: C 6 11 → B 5 11 + ? S 16 35 → ? + e − 1 0 ? → Y 39 90 + e − 1 0 B 83 210 i → ? + H 2 4 e ? → Y 39 89 + e + 1 0 A process in which interaction of two nuclear particles results in formation of one or more nuclear particles which are different from the parent nuclear particle is called nuclear reaction. Types of radioactive emission are: Alpha Emission: In this type of radioactive decay, an alpha particle (helium nucleus) is emitted by an atomic nucleus which results in an atom whose mass number is reduced by 4 and atomic number reduced by 2 . Beta Emission: In this type of radioactive decay, a beta ray is emitted from an atomic nucleus along with the neutrino. β (beta) particle is a fast-moving particle, symbol e − 1 0 is used to represent beta emission. Positron Emission: In this type of radioactive decay, a nucleus emits the positron ( e 1 0 ) from the nucleus of an atom. Gamma Emission: In this type of radioactive decay, a nucleus emits the gamma ( γ )-ray, which is high energy radiation. The type of radiation depends upon the nature of radiation like its mass number and charge. Formula used: A x m + B y n → C r p + D s q where A , B is reactant and C , D is product. m , n , p , q and x , y , r , s is the mass number and atomic number respectively. According to the basic principle of nuclear reaction: m + n = p + q and x + y = r + s . Calculation: The incomplete reaction is C 6 11 → B 5 11 + ? The mass number of the new nucleus is 11 = 11 + ? 11 – 11 = 0 The atomic number of the new nucleus is 6 = 5 + ? 6 – 5 = ? 6 – 5 = 1 The complete reaction is C 6 11 → B 5 11 + e + 1 0 ( positron emission ) The incomplete reaction is S 16 35 → ? + e − 1 0 The mass number of the new nucleus is 35 = ? + 0 35 – 0 = ?

Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

13th Edition

ISBN: 9780134421353

Author: Karen C. Timberlake

Publisher: PEARSON

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Question

Chapter 5.2, Problem 5.20PP

Interpretation Introduction

To determine:

Complete nuclear equations and type of radiations of given reactions.

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Chapter 5 Solutions

Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

Ch. 5.1 - Prob. 5.1PP Ch. 5.1 - Prob. 5.2PP Ch. 5.1 - Naturally occurring potassium consists of three...Ch. 5.1 - Naturally occurring iodine is iodine-127....Ch. 5.1 - Identify each of the following: a. 10X b. 24X c....Ch. 5.1 - Identify each of the following: a. 11X b. 3581X c....Ch. 5.1 - Prob. 5.7PP Ch. 5.1 - Prob. 5.8PP Ch. 5.1 - Prob. 5.9PP Ch. 5.1 - Prob. 5.10PP

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