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A nuclear transformation is given by $Y(n, \alpha) \rightarrow{ }_{3} \mathrm{Li}^{7}$. The nucleus of element $Y$ is
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${ }_{5}^{\mathrm{B}^{10}}$
$\mathrm{Y}(\mathrm{n}, \alpha)$ the nucleus splits into $\alpha$-particle and neutrons
i.e. $\mathrm{Z}^{\mathrm{Y}}{ }^{\mathrm{A}}+{ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{3} \mathrm{~L}_{\mathrm{i}}^{7}+{ }_{2} \mathrm{He}^{4}$
So $\mathrm{A}+1=7+4 \Rightarrow \mathrm{A}=10$
and $Z+0=3+2$ or $Z=5$
Hence, the nucleus of element $Y$ is boron ${ }_{5} \mathrm{Y}^{10}={ }_{5} \mathrm{~B}^{10}$
i.e. $\mathrm{Z}^{\mathrm{Y}}{ }^{\mathrm{A}}+{ }_{0} \mathrm{n}^{1} \longrightarrow{ }_{3} \mathrm{~L}_{\mathrm{i}}^{7}+{ }_{2} \mathrm{He}^{4}$
So $\mathrm{A}+1=7+4 \Rightarrow \mathrm{A}=10$
and $Z+0=3+2$ or $Z=5$
Hence, the nucleus of element $Y$ is boron ${ }_{5} \mathrm{Y}^{10}={ }_{5} \mathrm{~B}^{10}$
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