Q. In the nuclear reaction: $X(n,\alpha)\ _{3}Li^{7}$ the term $X$ will be
A) $\ _{5}B^{10}$
B) $\ _{5}B^{9}$
C) $\ _{5}B^{11}$
D) $\ _{2}He^{4}$
Explanation
(a) $X(n,\alpha)\ _{3}^{7}Li \Rightarrow \ _{Z}X^{A} + \ _{0}n^{1} \rightarrow 3^{Li^{7}} +_{2}He^{4}$
$Z = 3 + 2 = 5$ and $A = 7 + 4 - 1 = 10$
$\therefore_{5}X^{10} =_{5}B^{10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.