Bombardment of aluminum by alpha-particle lea | vedclass

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(A) The reaction for path $(i)$ is: ${ }_{13}^{27} Al +{ }_2^4 \alpha \rightarrow{ }_{14}^{30} Si +{ }_1^1 p (X)$.Here,$X$ is a proton $({ }_1^1 p)$ b

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proton,neutron,positron

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(A) The reaction for path $(i)$ is: ${ }_{13}^{27} Al +{ }_2^4 \alpha \rightarrow{ }_{14}^{30} Si +{ }_1^1 p (X)$.Here,$X$ is a proton $({ }_1^1 p)$ because the mass number and atomic number are conserved ($27+4 = 30+1$ and $13+2 = 14+1$).The reaction for path $(ii)$ is: ${ }_{13}^{27} Al +{ }_2^4 \alpha \rightarrow{ }_{15}^{30} P +{ }_0^1 n (Y)$.Here,$Y$ is a neutron $({ }_0^1 n)$ because the mass number and atomic number are conserved ($27+4 = 30+1$ and $13+2 = 15+0$).The subsequent decay of phosphorus is: ${ }_{15}^{30} P \rightarrow{ }_{14}^{30} Si +{ }_{+1}^{0} e (Z)$.Here,$Z$ is a positron $({ }_{+1}^{0} e)$ because the atomic number decreases by $1$.Therefore,$X$ is proton,$Y$ is neutron,and $Z$ is positron.Thus,the correct option is $A$.

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

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