Question

Match the following reactions:

$\left(A\right)$ ${}_4{Be}^9{+}_2{He}^4{\to }_6{C}^{12}+\cdots$	$\left(i\right)$ ${}_2{He}^4$
$\left(B\right)$ ${}_6{C}^{12}+\cdots {\to }_5{B}^{10}{+}_2{He}^4$	$\left(ii\right)$ ${}_0{n}^1$
$\left(C\right)$ ${}_7{N}^{14}+\cdots {\to }_8{O}^{17}{+}_1{H}^1$	$\left(iii\right)$ ${}_1{D}^2$
$\left(D\right)$ ${}_{20}{Ca}^{40}+\cdots {\to }_{19}{K}^{37}{+}_2{He}^4$	$\left(iv\right)$ ${}_1{H}^1$

• A. $A-i,B-ii,C-iii,D-iv$
• B. $A-ii,B-iii,C-i,D-iv$
• C. $A-iv,B-ii,C-iii,D-i$
• D. $A-iii,B-ii,C-i,D-iv$

Answer 1

The correct option is B $A-ii,B-iii,C-i,D-iv$

$4^{{Be}^9}{+}_2^{4}He{⟶}_6^{12}C{+}_0^{1}n$ (neutron emission)

$6^{C^{12}}+1^{D^2}{⟶}_5^{10}B+2^{{He}^4}$

$7^{N^{14}}{+}_2^{4}He⟶8^{O^{17}}{+}_1^{1}H$

${20}^{{Ca}^{40}}{+}_1^{1}H⟶{19}^{K^{37}}+2^{{He}^4}$