Bombardment of aluminum by an $$\alpha$$-particle leads to its artificial disintegration in two ways, as shown in the reactions below. The resultant products X, Y, and Z are identified as follows :

Reaction (i)

Disintegration of aluminum into silicon :

$$^{4}_{2} \text{He} + ^{27}_{13} \text{Al} \rightarrow ^{30}_{14} \text{Si} + ^{A}_{Z} X$$

Conservation conditions :

(a) Charge balance :

$ 2 + 13 = 14 + Z \implies Z = 1 $

(b) Mass balance :

$ 4 + 27 = 30 + A \implies A = 1 $

The particle $A_Z X$ is $^1_1 \text{H}$, a proton (an isotope of hydrogen).

Reaction (ii)

Disintegration of aluminum into phosphorus :

$$^{4}_{2} \text{He} + ^{27}_{13} \text{Al} \rightarrow ^{30}_{15} \text{P} + ^{A}_{Z}Y$$

Conservation conditions :

(a) Charge balance :

$ 2 + 13 = 15 + Z \implies Z = 0 $

(b) Mass balance :

$ 4 + 27 = 30 + A \implies A = 1 $

The particle $Y$ has one unit mass and no charge, which is a neutron $^1_0 n$.

Reaction (iii)

Disintegration of phosphorus into silicon :

$$^{30}_{15}\text{P} \longrightarrow \, ^{30}_{14}\text{Si} + ^{A}_{Z}\text{Z} $$

Conservation conditions :

(a) Mass balance :

$ 30 = 30 + A \implies A = 0 $

(b) Charge balance :

$ 15 = 14 + Z \implies Z = 1 $

The particle has zero mass but one unit of positive charge. Hence, the particle is a positron $\left(^{0}_{+1}\beta\right)$.

Summary

The particles are :

X = $^1_1 \text{H}$ (proton)

Y = $^1_0 n$ (neutron)

Z = $^{0}_{+1}\beta$ (positron)