Bombardment of aluminium be -particle leads to its artificial disintegration in two way (i) and (ii) — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Bombardment of aluminium be $\alpha$-particle leads to its artificial disintegration in two way (i) and (ii) as shown. Products X, Y and Z, respectively, are <br> ${}_{13}^{27}\text{Al} \xrightarrow{\text{(i)}} {}_{14}^{30}\text{Si} + \text{X}$ <br> ${}_{13}^{27}\text{Al} \xrightarrow{\text{(ii)}} {}_{15}^{30}\text{P} + \text{Y}$ <br> ${}_{15}^{30}\text{P} \xrightarrow{\quad} {}_{14}^{30}\text{Si} + \text{Z}$
💡 Solution & Explanation
The initial reacting complex is ${}_{13}^{27}\text{Al} + {}_2^4\text{He}$ (total mass 31, Z=15). In path (i): $31=30+A_X \implies A_X=1$; $15=14+Z_X \implies Z_X=1$ (X is a proton). In path (ii): $31=30+A_Y \implies A_Y=1$; $15=15+Z_Y \implies Z_Y=0$ (Y is a neutron). Decay of P-30: $30=30+A_Z \implies A_Z=0$; $15=14+Z_Z \implies Z_Z=1$ (Z is a positron). Therefore, correct answer is A.