Chapter 31, Problem 2Q

To determine

The resulting nuclide.

Answer to Problem 2Q

Solution:

The resulting nuclide is neon and the reaction equation is, ${}_{11}{}^{22}Na+{}_1{}^{2}H\to {}_{10}{}^{20}Ne+{}_2{}^4\alpha$

Explanation of Solution

Given:

${}_{11}{}^{22}Na$ is bombarded with deuteron produces an alpha particle and another nuclide.

The reaction equation is,

${}_{11}{}^{22}Na+{}_1{}^{2}H\to {}_A{}^{B}X+{}_2{}^4\alpha$

From the law of conservation of nucleons, there will be two equations.

$\begin{array}{l}22+2=B+4------(1)\\ 11+1=A+2------(2)\end{array}$

From these,

$\begin{array}{l}A=10\\ B=20\end{array}$

The atomic number 10 corresponds to the neon atom.