Question

The following disintegration reactions are
(i) ${}_z{X^A}{\text{ ?}} \to {\text{ }}{}_{z - 2}{Y^A}{\text{ + ?}} {}_z{X^A}{\text{ ?}} \to {\text{ }}{}_{z - 2}{Y^A}{\text{ + ?}}$
(ii) ${}_z{X^A}{\text{ ?}} \to {\text{ }}{}_{z - 1}{P^A}{\text{ + ?}}$

Answer 1

Hint:First we have to know the atomic and mass number of alpha particles $\alpha $ which is $2{\text{ and 4}}$ respectively. For beta emission $\beta $, there is an increase in atomic number by 1 and mass remains unaffected. Thus from the questions by seeing the difference in the mass numbers and atomic numbers we can determine whether it's an alpha emission or beta emission. There’s another type of emission known as gamma emission which is involved with energy and has no indulgence with atomic and mass number.

Complete answer:
(i) We see in the first equation that there’s a difference of 2 in the atomic number, hence it can be an alpha emission but when we see the mass number, we see it remains the same, without decreasing by $4$. Thus we can surely say that the question given is wrong.

(ii) In the second case we see that there’s no change in the mass number , thus seeing this we can guess that it is a beta emission, but looking at the atomic number we see that it has not increased by one, hence it cannot be a beta emission, nor an alpha emission, hence this equation too has been wrong.

Note: For questions such as these, always remember that for alpha emission there is a loss of 2 in atomic number and a loss of 4 in mass number and for beta emission there is no change in the mass number but there is an increase of one in the atomic number. It is very important to remember that there is an increase in atomic number in beta emission and not decrease.