Question

A certain radioactive material $_{Z} X ^{A}$ starts emitting $\alpha$ and $\beta$ particles successively such that the end product is $_{z-3} Y ^{A-8}$. The number of $\alpha$ and $\beta$ particles emitted are

• A. 4 and 3 respectively
• B. 2 and 1 respectively
• C. 3 and 4 respectively
• D. 3 and 8 respectively

Answer 1

Answer: (B)

2 and 1 respectively

Answer 2

Let there be $x \alpha$-particles and $y\,\beta$-particles ${ }_{z} X^{A} \longrightarrow x He _{2}^{4}+y \beta_{-1}^{0}+Y_{Z-3}^{A-8}$ then equating the mass numbers $A=4 x+A-8\,\,\,...(i)$ and equating atomic numbers $Z=2 x-y+Z-3\,\,\,...(ii)$ Solving Eqs. (i) and (ii), we get $x=2$ and $y=1$ $\therefore$ The number of $\alpha$ and $\beta$ particles emitted are $2$ and $1$ respectively.