a proton
a gamma ray
a helium particle
an alpha particle
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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