Chapter 31, Problem 4P

To determine

Whether the reaction $\mathrm{p}{+}_3^7\text{Li}{\to }_2^4\text{He}+\mathrm{\alpha }$ requires energy or does it release the energy and also to determine the amount of energy released.

Answer to Problem 4P

The release in the energy is of $17.35\text{MeV}$ .

Explanation of Solution

Given Information:

The reaction is $\mathrm{p}{+}_3^7\text{Li}{\to }_2^4\text{He}+\mathrm{\alpha }$ .

Calculation:

Consider that in the reaction ${}_3^7\text{Li}\left(\mathrm{P},\mathrm{\alpha }\right)\underset{\text{2}}{\overset{\text{4}}{\text{He}}}\underset{\text{1}}{\overset{\text{1}}{\text{H}}}\text{+}\underset{\text{3}}{\overset{\text{7}}{\text{Li}}}\to \underset{\text{2}}{\overset{\text{4}}{\text{He}}}\text{+}\underset{\text{2}}{\overset{\text{4}}{\text{He}}}$ .

Consider the mass of $0\left(2\times \underset{2}{\overset{4}{\mathrm{H}\mathrm{e}}}\right)$ is $2\times 4.002603\text{u}$ . Then, the mass is ${\mathrm{m}}_1=8.005206\text{u}$ .

Consider the mass,

  $\begin{array}{l}\left(\underset{1}{\overset{1}{\mathrm{H}}}+\underset{3}{\overset{7}{\mathrm{L}\mathrm{i}}}\right)=\left(1.007825+7.01600\right)\mathrm{u}\\ {\mathrm{m}}_2=8.023829\text{u}\end{array}$

Consider the mass defect is,

  $\begin{array}{c}\mathrm{\Delta }\mathrm{m}=\left({\mathrm{m}}_2-{\mathrm{m}}_1\right)\\ =0.018623\text{u}\end{array}$

Then, the release in the energy is,

  $\left(0.018623\text{u}\right)\left(\frac{931.5\text{MeV}/{\text{c}}^2}{\mathrm{u}}\right){\mathrm{c}}^2=17.35\text{MeV}$