Q. A nucleus $_{n}^{m } X ⁡$ emits one α particle and two $\beta ^{-}$ particles. The resulting nucleus is

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A

$_{n - 4}^{m ⁡ - 6} Z ⁡$

B

$_{n}^{m ⁡ - 6} Z ⁡$

C

$_{n}^{m - 4} X ⁡$

D

$_{n - 2}^{m ⁡ - 4} Y ⁡$

Solution:

When an alpha particle $\left(_{2}^{4} \text{He}\right)$ is emitted, the mass number and the atomic number of the daughter nucleus decreases by four and two respectively. When a beta particle (β^–) is emitted, the atomic number of the daughter nucleus increases by one but the mass number remains the same.
$∴ _{n}^{m } X ⁡ \overset{\alpha }{ \rightarrow } _{n - 2}^{m ⁡ - 4} Y ⁡ \overset{2 \beta ^{-}}{ \rightarrow } _{n}^{m ⁡ - 4} X ⁡$

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