Chapter 24.2, Problem 15SSC

Interpretation Introduction

Interpretation:

The nuclear equation for the alpha decay of radium-226 is to be written.

Concept introduction:

Nuclear reactions are expressed by balanced nuclear reactions just as chemical reactions are expressed by balanced nuclear reactions. However, in balanced chemical equations, numbers and type of atoms are conserved but in nuclear equations, mass numbers and charges are conserved.

Answer to Problem 15SSC

The nuclear equation for the alpha decay of radium-226 is,

$\text{R}\text{a}\to \text{R}\text{n+}\text{H}\text{e}$

Explanation of Solution

In alpha decay, an alpha releases and decrease of mass number and atomic number takes place by 4 and 2 respectively. So, the alpha decay of radium-226 is as follows:

$\text{R}\text{a}\to \text{R}\text{n+}\text{H}\text{e}$

The above nuclear equation is balanced as the mass number in both sides is 226 and the atomic number is 88.

Conclusion

In alpha decay, atomic number decrease by 2 and mass number decreases by 4.