Question

In the nuclear reaction given by ${}_{2}H{e}^4{+}_7{N}^{14}\to {}_1{H}^1+X$ then the nucleus $X$ is :

• A. Nitrogen of mass 16
• B. Nitrogen of mass 17
• C. Oxygen of mass 16
• D. Oxygen of mass 17

Answer 1

The correct option is D Oxygen of mass 17

Let the nucleus $X$ has mass number $A$ and atomic number $Z$ :
Conservation of atomic number :
$\therefore$ $2+7=1+Z$
$⟹Z=8$
Conservation of mass number :
$\therefore$ $4+14=1+A$
$⟹A=17$
Thus the nucleus $X$ is ${}_8{O}^{17}$.