Question

In the nuclear reaction -

${}_7{}^{14}\mathrm{N}+{}_2{}^4\mathrm{He}\to {}_8{}^{17}\mathrm{O}+........$

• A. ${}_2{}^4\mathrm{He}$
• B. ${}_1{}^1\mathrm{H}$
• C. ${}_0{}^1\mathrm{n}$
• D. $\mathrm{\gamma }$

Answer 1

The correct option is B

${}_1{}^1\mathrm{H}$

The nuclear reaction taking place is:

${}_7{}^{14}\mathrm{N}+{}_2{}^4\mathrm{He}\to {}_8{}^{17}\mathrm{O}+........$

Consider the blank space is $X$.

Equate the mass number of both sides of the reaction.

$\begin{array}{rcl}14+4& =& 17+X\\ X& =& 18-17\\ X& =& 1\end{array}$

Equate the atomic mass of both sides of the reaction.

$\begin{array}{rcl}7+2& =& 8+X\\ X& =& 9-8\\ X& =& 1\end{array}$

It can be concluded that $X$ has atomic number 1 and mass number 1. So, the blank element is ${}_1{}^1\mathrm{H}$.

Hence, option (B) is correct.