Question

A nuclear transformation is given by $Y(n,\alpha ){\to }_{3}L{i}^7$. The nucleus of element $Y$ is

• A. ${}_{5}B{e}^{11}$
• B. ${}_5{B}^{10}$
• C. ${}_5{B}^9$
• D. ${}_6{C}^{12}$

Answer 1

Answer: (B)

${}_5{B}^{10}$

$Y(n,\alpha )$ the nucleus splits into $\alpha$-particle and neutrons
i.e. ${}_z{Y}^A{+}_0{n}^1{\to }_{3}L{i}^7{+}_{2}H{e}^4$
So, $A+1=7+4\Rightarrow A=10$
and $Z+0=3+2$ or $Z=5$
Hence, the nucleus of element Y is boron ${}_5{Y}^{10}{=}_5{B}^{10}$.