Question

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

• A. ${}_6{C}^{12}$
• B. ${}_5{B}^{10}$
• C. ${}_5{B}^9$
• D. ${}_{4}B{e}^{11}$

Answer 1

The correct option is B ${}_5{B}^{10}$

The notation means that on being bombarded with neutrons the element breaks into Lithium and an $\alpha -particle.$
${}_A^{Z}X{+}_0^{1}n{\Rightarrow }_{A-2}^{Z-3}Y{+}_2^{4}He$

Now,
${}_{A-2}^{Z-3}Y{=}_3^{7}Li$

$\therefore Z=10,A=5$
So, The element is Boron and X is ${}_5{B}^{10}$