Chapter 31, Problem 27P

To determine

To proof:The energy released in the fusion reaction is 3.27 MeV.

Explanation of Solution

Given Information:

The energy from ${}_1^2\text{H}{+}_1^2\text{H}{\to }_2^3\text{He}+\text{n}$ is 3.27 MeV.

Formula used:

Consider the expression to calculate the required value.

  $\mathrm{Q}=\left({\mathrm{m}}_{{}_1^2\text{H}}+{\mathrm{m}}_{{}_1^2\text{H}}-{\mathrm{m}}_{{}_2^3\text{He}}-{\text{m}}_{\text{n}}\right){\mathrm{c}}^2$

Calculation:

Substitute the values in the above equation.

  $\begin{array}{c}\mathrm{Q}=\left({\mathrm{m}}_{{}_1^2\text{H}}+{\mathrm{m}}_{{}_1^2\text{H}}-{\mathrm{m}}_{{}_2^3\text{He}}-{\text{m}}_{\text{n}}\right){\mathrm{c}}^2\\ =\left(2.014102+2.014102-3.016029-1.008665\right)\times 93.15\\ =3.27\text{ MeV}\end{array}$

Conclusion:

Thus, the energy from ${}_1^2\text{H}{+}_1^2\text{H}{\to }_2^3\text{He}+\text{n}$ is 3.27 MeV.