Question

On the bombardment of Boron with neutron, an $\alpha$ - particle is emitted and product nucleus formed is $\underset{–}{}$

• A. ${}_6{C}^{12}$
• B. ${}_{2}L{i}^6$
• C. ${}_{3}L{i}^7$
• D. ${}_{4}B{e}^9$

Answer 1

Answer: (C)

${}_{3}L{i}^7$

${}_5{B}^{10}{+}^{1}n{\to }_Z^{A}X{+}_2^{4}He$
Mass balance: $10+1=A+4$
Atomic no balance: $5+0=Z+2$
$Z=3$
Therefore, it's ${}_{3}L{i}^7$