\(^{27}_{13}Al\) + \(^{1}_{0}n\) \(\Rightarrow\) \(^{24}_{11}Na\) + X
- a proton
- a gamma ray
- a helium particle
-
an alpha particle
The correct answer is: D
Explanation
\(^{27}_{13}Al\) + \(^{1}_{0}n\) \(\Rightarrow\) \(^{24}_{11}Na\) + \(^{b}_{a}X\)a + 11 = 0 + 13 \(\Rightarrow\) a = 2
b + 24 = 1 + 27 \(\Rightarrow\) b = 4
\(^{b}_{a}X\) = \(^{4}_{2}\alpha\)