Chapter 8, Problem 8.1YT

Interpretation Introduction

Interpretation:

The nuclear equation that represents alpha decay is to be written.

Concept Introduction:

An alpha particle consists of two protons and two neutrons. It is a He nucleus emitted by a decaying atomic nucleus. Alpha particles can be represented with the symbol, ${{}^4}_{\text{2}}{\text{He}}^{}$.

Answer to Problem 8.1YT

Solution:

${{}^{226}}_{88}\text{Ra}{\to }^{222}{}_{86}\text{Rn}+\text{H}\text{e}+\gamma$

Explanation of Solution

An alpha particle consists of two protons and two neutrons. During alpha decay, reactant reacts and emits an alpha particle and attains an atomic number that is two less than its original atomic number and a mass number that is four less than its original mass number.

Thus, Ra-226 decays through alpha particle emission to form ${{}^{222}}_{86}\text{Rn}$.

It is to be noted that the sum of the numbers present in superscripts, that is the mass numbers of the reactants and the products, should be the same on both side. Also, the sums of number present in subscripts, that is the atomic numbers, should be the same on both sides.

Conclusion

In this reaction, ${{}^{226}}_{88}\text{Ra}$ decays via alpha emission to produce ${{}^{222}}_{86}\text{Rn}$. The equation is written as:

${{}^{226}}_{88}\text{Ra}{\to }^{222}{}_{86}\text{Rn}+\text{H}\text{e}+\gamma$