Question

In the nuclear fission reaction, ${}_2^{4}He+{}_7^{14}N\to {}_p^{q}x+{}_1{H}^1$ the nucleus ${}_p^{q}X$ is :

• A. nitrogen of mass 16
• B. nitrogen of mass 17
• C. oxygen of mass 16
• D. oxygen of mass 17

Answer 1

Answer: (D)

In order that reaction holds true mass number and atomic number remain conserved.
Given ${}_2^{4}He+{}_7^{14}N\to {}_p^{q}x+{}_1{H}^1$
Equating mass number $14+4=q+1$
$\Rightarrow$ $q=17$
Equating atomic number $7+2=p+1$
$\Rightarrow$ $p=8$
Hence, unknown element is an isotope of oxygen of mass $17$.