Chapter 31, Problem 10P

To determine

The total kinetic energy of the products of the reaction $\mathrm{d}{+}_6^{13}\text{C}{\to }_7^{14}\text{N}+\text{n}$ if the incoming (d) has the kinetic energy 36.3 MeV.

Answer to Problem 10P

The total kinetic energy is $41.6\text{MeV}$ .

Explanation of Solution

Given Information:

The kinetic energy is 36.3 MeV.

Calculation Consider the reaction is,

  $\underset{1}{\overset{2}{\mathrm{H}}}{+}_6^{13}\text{C}{\to }_7^{14}\text{N}+\underset{0}{\overset{1}{\mathrm{n}}}$

Consider the mass.

  $\begin{array}{l}\left({}_7^{14}\text{N}+\underset{0}{\overset{1}{\mathrm{n}}}\right)=\left(14.003074+1.008665\right)\text{u}\\ {\mathrm{m}}_1=15.011739\text{u}\end{array}$

Consider the mass is,

  $\begin{array}{l}\left({}_1^2\mathrm{H}+\underset{6}{\overset{13}{\mathrm{C}}}\right)=\left(2.014102+13.003355\right)\mathrm{u}\\ {\mathrm{m}}_2=15.017457\text{u}\end{array}$

Then, the defect mass is,

  $\mathrm{\Delta }\mathrm{m}=+0.005718\text{u}$

The energy released is,

  $\left(0.005718\times 931.5\right)\text{MeV}=+\text{5}\text{.326317}\text{MeV}$

The total kinetic energy is,

  $\begin{array}{c}\left(36.3+\text{5}\text{.326317}\right)\text{MeV}=\text{41}\text{.626317}\text{MeV}\\ =41.6\text{MeV}\end{array}$