Question

In the nuclear reaction, ${}_7^{14}N+X{⟶}_6^{14}C{+}_1^{1}H$, the $X$ will be

• A. ${}_{-1}^{0}e$
• B. ${}_1^{2}H$
• C. ${}_1^{1}H$
• D. ${}_0^{1}n$

Answer 1

Answer: (D)

${}_0^{1}n$

Let $X$ has mass number and atomic number as $A$ and $Z$ respectively.

In nuclear reaction: ${}_7^{14}N{+}_Z^{A}X\to$ ${}_6^{14}C{+}_1^{1}H$

Applying conservation of mass number :

$\Rightarrow 14+A=14+1$

$\Rightarrow$ $A=1$

Applying conservation of atomic number :

$\Rightarrow 7+Z=6+1$

$\Rightarrow$ $Z=0$

Thus the element $X$ is neutron, i.e, ${}_Z^{A}X{=}_0^{1}n$