Question

Bombardment of aluminum by $\alpha$-particle leads to its artificial disintegration in two ways, (i) and (ii) as shown in the given figure. Products $X,Y$ and $Z$ respectively are:

• A. proton, neutron, positron
• B. neutron, positron, proton
• C. proton, positron, neutron
• D. positron, proton, neutron

Answer 1

Answer: (A)

proton, neutron, positron

a) ${}_{13}^{27}Al{+}_2^4\alpha {\to }_{14}^{30}Si{+}_1^{1}P\left(X\right)$

Here, X is proton since this leads to conservation of mass number $(27+4=30+1)$ and atomic number $(13+2=14+1)$ in the reaction.

b) ${}_{13}^{27}Al{+}_2^4\alpha {\to }_{15}^{30}P{+}_0^{1}n\left(Y\right)$

Here, Y is neutron since this leads to conservation of mass number $(27+4=30+1)$ and atomic number $(13+2=15+0)$ in the reaction.

c) ${}_{15}^{30}P{\to }_{14}^{30}Si+{}_{+1}^{0}B\left(Z\right)$

Z is positron since there is decrease in atomic number by 1 in the reaction.

Option A is correct.