Question

A radioactive atom ${}_Y^{X}M$ emits two $\alpha$ particles and one $\beta$ particle successively. The number of neutrons in the nucleus of the product will be :

• A. $X-4-Y$
• B. $X-Y-5$
• C. $X-Y-3$
• D. $X-Y-6$

Answer 1

The correct option is B $X-Y-5$

An $\alpha$-emission reduces the atomic mass by $4$ and atomic number by $2$, whereas a $\beta$-emission increases the atomic number by $1$ with no change in atomic mass.

$M_Y^X\underset{}{\overset{-\alpha }{\to }}{N}_{Y-2}^{X-4}\underset{}{\overset{-\alpha }{\to }}{O}_{Y-4}^{X-8}\underset{}{\overset{-\beta }{\to }}{P}_{Y-3}^{X-8}$

Number of neutrons $=$ Mass number $-$ Atomic number

$=X-8-Y+3$

$=X-Y-5$

Hence, option B is correct.