$^{27}_{13}Al$ + $^{1}_{0}n$ $\Rightarrow$ $^{24}_{11}Na$ + X

a proton

a gamma ray

a helium particle

an alpha particle

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Correct answer is D

$^{27}_{13}Al$ + $^{1}_{0}n$ $\Rightarrow$ $^{24}_{11}Na$ + $^{b}_{a}X$

a + 11 = 0 + 13 $\Rightarrow$ a = 2

b + 24 = 1 + 27 $\Rightarrow$ b = 4

$^{b}_{a}X$ = $^{4}_{2}\alpha$

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