Chapter 31, Problem 28P

To determine

To proof:The energy released in the fusion reactionis 12.86 MeV.

Explanation of Solution

Given Information:

The energy from ${}_1^1\text{H}{+}_1^1\text{H}{\to }_1^2\text{H}+{\text{e}}^{+}+\mathrm{v}$ is 0.42 MeV

The energy from ${}_1^1\text{H}{+}_1^2\text{H}{\to }_1^3\text{He}+\mathrm{\gamma }$ is 5.49 MeV

The energy from ${}_2^3\text{H}{+}_2^3\text{H}{\to }_2^4\text{He}{+}_1^1\text{H}{+}_1^1\text{H}$ is 12.86 MeV

Formula used:

Calculate the required value.

  $\mathrm{Q}=\left({\mathrm{m}}_{{}_1^1\text{H}}+{\mathrm{m}}_{{}_1^1\text{H}}-{\mathrm{m}}_{{}_1^2\text{H}}-{\text{m}}_{{\text{e}}^{+}}-{\mathrm{m}}_{{\mathrm{e}}^{-}}\right){\mathrm{c}}^2$  ... (1)

Calculation:

Substitute the values in the above equation.

  $\begin{array}{c}\mathrm{Q}=\left({\mathrm{m}}_{{}_1^1\text{H}}+{\mathrm{m}}_{{}_1^1\text{H}}-{\mathrm{m}}_{{}_1^2\text{H}}-{\text{m}}_{{\text{e}}^{+}}-{\mathrm{m}}_{{\mathrm{e}}^{-}}\right){\mathrm{c}}^2\\ =\left(1.007825+1.007825-2.014102-0.000549-0.000549\right)\times 93.15\\ =0.42\text{ MeV}\end{array}$

Calculate the required value.

  $\mathrm{Q}=\left({\mathrm{m}}_{{}_1^1\text{H}}+{\mathrm{m}}_{{}_1^2\text{H}}-{\mathrm{m}}_{{}_2^3\text{He}}-{\text{m}}_{\mathrm{\gamma }}\right){\mathrm{c}}^2$   ... ( (2)

Substitute the values in the above equation.

  $\begin{array}{c}\mathrm{Q}=\left({\mathrm{m}}_{{}_1^1\text{H}}+{\mathrm{m}}_{{}_1^2\text{H}}-{\mathrm{m}}_{{}_2^3\text{He}}-{\text{m}}_{\mathrm{\gamma }}\right){\mathrm{c}}^2\\ =\left(1.007825+2.014102-3.016029-0\right)\times 93.15\\ =5.49\text{ MeV}\end{array}$

Calculate the required value.

  $\mathrm{Q}=\left({\mathrm{m}}_{{}_2^3\text{H}}+{\mathrm{m}}_{{}_2^3\text{H}}-{\mathrm{m}}_{{}_3^4\text{He}}-{\mathrm{m}}_{{}_1^1\text{H}}-{\mathrm{m}}_{{}_1^1\text{H}}\right){\mathrm{c}}^2$   ... ( (3)

Substitute the values in the above equation.

  $\begin{array}{c}\mathrm{Q}=\left({\mathrm{m}}_{{}_2^3\text{H}}+{\mathrm{m}}_{{}_2^3\text{H}}-{\mathrm{m}}_{{}_3^4\text{He}}-{\mathrm{m}}_{{}_1^1\text{H}}-{\mathrm{m}}_{{}_1^1\text{H}}\right){\mathrm{c}}^2\\ =\left(3.016029+3.016029-4.002603-1.007825-1.007825\right)\times 93.15\\ =12.86\text{ MeV}\end{array}$

Conclusion:

Thus, the energy from ${}_1^1\text{H}{+}_1^1\text{H}{\to }_1^2\text{H}+{\text{e}}^{+}+\mathrm{v}$ is 0.42 MeV, the energy from ${}_1^1\text{H}{+}_1^2\text{H}{\to }_1^3\text{He}+\mathrm{\gamma }$ is 5.49 MeV, the energy from ${}_2^3\text{H}{+}_2^3\text{H}{\to }_2^4\text{He}{+}_1^1\text{H}{+}_1^1\text{H}$ is 12.86 MeV.