When a deuteron is bombarded on ${ }_8^{16} O$ nucleus, an $\alpha$-particle is emitted. Other product nucleus is:

${ }_7^{14} N$

The correct answer is Option (4) → ${ }_7^{14} N$

The reaction can be written as -

${ }_8^{16} O+{ }_1^{2}H→{ }_2^{4}He+X$

According to conservation of Mass,

$16+2=4+m_x$

$m_x=14$

and, conservation of atomic number

$8+1=2+Z_x$

$Z_x=7$

$∴X={ }_7^{14} N$