Artificial Disintegration and Products of Aluminum Bombardment

To determine the products $\mathbf{X}$, $\mathbf{Y}$, and $\mathbf{Z}$, we apply the principles of conservation of mass number ($A$) and atomic number ($Z$) for each nuclear reaction.

An $\alpha$-particle is represented as ${}_{2}^{4}\text{He}$.

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1. Path (i): Formation of $\mathbf{X}$

The nuclear equation for path (i) is:
${}_{13}^{27}\text{Al} + {}_{2}^{4}\text{He} \longrightarrow {}_{14}^{30}\text{Si} + \mathbf{X}$

• Conservation of Mass Number ($A$):
  $27 + 4 = 30 + A_{\mathbf{X}} \implies A_{\mathbf{X}} = 1$

• Conservation of Atomic Number ($Z$):
  $13 + 2 = 14 + Z_{\mathbf{X}} \implies Z_{\mathbf{X}} = 1$

Since $\mathbf{X}$ has $A = 1$ and $Z = 1$, it is a proton (${}_{1}^{1}\text{p}$ or ${}_{1}^{1}\text{H}$).

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2. Path (ii): Formation of $\mathbf{Y}$

The nuclear equation for path (ii) is:
${}_{13}^{27}\text{Al} + {}_{2}^{4}\text{He} \longrightarrow {}_{15}^{30}\text{P} + \mathbf{Y}$

• Conservation of Mass Number ($A$):
  $27 + 4 = 30 + A_{\mathbf{Y}} \implies A_{\mathbf{Y}} = 1$

• Conservation of Atomic Number ($Z$):
  $13 + 2 = 15 + Z_{\mathbf{Y}} \implies Z_{\mathbf{Y}} = 0$

Since $\mathbf{Y}$ has $A = 1$ and $Z = 0$, it is a neutron (${}_{0}^{1}\text{n}$).

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3. Decay of ${}_{15}^{30}\text{P}$: Formation of $\mathbf{Z}$

The nuclear equation for the disintegration of phosphorus-30 is:
${}_{15}^{30}\text{P} \longrightarrow {}_{14}^{30}\text{Si} + \mathbf{Z}$

• Conservation of Mass Number ($A$):
  $30 = 30 + A_{\mathbf{Z}} \implies A_{\mathbf{Z}} = 0$

• Conservation of Atomic Number ($Z$):
  $15 = 14 + Z_{\mathbf{Z}} \implies Z_{\mathbf{Z}} = 1$

Since $\mathbf{Z}$ has $A = 0$ and $Z = +1$, it is a positron (${}_{+1}^{0}\text{e}$ or $\text{e}^+$).

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Conclusion:

The products $\mathbf{X}$, $\mathbf{Y}$, and $\mathbf{Z}$ are proton, neutron, and positron respectively.

Correct Option: A