Question

A nucleus $_n{X^m}$ emits one $\alpha $ and one $\beta $ particle. The resulting nucleus is
A. $_n{X^{m - 4}}$
B. $_{n - 2}{X^{m - 4}}$
C. $_{n - 4}{X^{m - 4}}$
D. $_{n - 1}{X^{m - 4}}$

Answer 1

Hint:In order to solve this question, we will first find the formed new nucleus due to an alpha decay using alpha decay rule and then with help of newly formed nucleus we will apply the beta decay rule to find final nucleus and then we will determine the correct option.

Formula used:
A general element is represented as $_x{Y^z}$, where Y is a general element, x is atomic number and z is atomic mass. An alpha decay result reduces the atomic number by two and atomic mass by four while a beta decay increases the atomic number by one.

Complete step by step solution:
According to the question, we have given that we have a nucleus X having atomic number n, atomic mass m $_n{X^m}$ and firstly it goes through alpha decay and as we know alpha decay will reduce the atomic number by two and atomic mass by four, so whole process can be represented as,
$_n{X^m}{\mathop \longrightarrow \limits^{\alpha - decay} _{n - 2}}{X^{m - 4}} \\ $
Now, this newly formed element X will go through a beta decay and beta decay will increase the atomic number by one and this process can be represented as
$_{n - 2}{X^{m - 4}}{\mathop \longrightarrow \limits^{\alpha - decay} _{n - 1}}{X^{m - 4}}$
So, the resultant nucleus is $_{n - 1}{X^{m - 4}}$.

Hence, the correct answer is option D.

Note: While solving such radioactive decay processes, firstly always write every elements with their atomic number and atomic mass in standard form and also remember the rule of alpha decay, beta decay and gamma decay to avoid any calculation mistakes. . and additionally in gamma decay there is no change in atomic number and in atomic mass as well, nucleus just release energy.