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X(n, $\alpha$) ${}^{7}_{3}Li$, then X will be : -
A
${}^{10}_{5}B$
B
${}^{9}_{5}B$
C
${}^{11}_{4}Be$
D
${}^{4}_{2}He$
Detailed Solution
X(n, $\alpha$)Li means: X absorbs a neutron and emits an $\alpha$-particle, leaving ${}^{7}_{3}Li$.
${}^{A}_{Z}X + {}^{1}_{0}n \rightarrow {}^{7}_{3}Li + {}^{4}_{2}He$
Mass number: $A + 1 = 7 + 4 \Rightarrow A = 10$
Atomic number: $Z + 0 = 3 + 2 \Rightarrow Z = 5$
Z = 5 is boron, so X is ${}^{10}_{5}B$.
${}^{A}_{Z}X + {}^{1}_{0}n \rightarrow {}^{7}_{3}Li + {}^{4}_{2}He$
Mass number: $A + 1 = 7 + 4 \Rightarrow A = 10$
Atomic number: $Z + 0 = 3 + 2 \Rightarrow Z = 5$
Z = 5 is boron, so X is ${}^{10}_{5}B$.