Question

Calculate the amount of energy released in kilojoules per gram of deuterium (${}^{\mathbf{2}}\mathbf{H}$), for the fusion reaction

${}_{\mathbf{1}}{}^{\mathbf{2}}\mathbf{H}\mathbf{+}{}_{\mathbf{1}}{}^{\mathbf{2}}\mathbf{H}\to {}_{\mathbf{2}}{}^{\mathbf{4}}\mathbf{H}$

Use the atomic masses in Table 19.1. Compare your answer with that of the preceding problem

Short answer

The amount of energy released per gram of deuterium in fission reaction is $5.6\times {10}^{8}KJ/g$.

Step-by-step solution

Calculating the mass difference

In a nuclear fusion reaction two lighter nuclides combine to form a heavier nuclide. The equation given below represents a nuclear fusion reaction.

$2{}_{\text{1}}{}^{\text{2}}\text{H}\to {}_{\text{2}}{}^{\text{4}}\text{H}$

We have to calculate the mass difference for the fusion reaction:

$\begin{array}{rcl}\Delta m& =& m\left({}_2{}^{4}He\right)-2m\left({}_1{}^{2}H\right)\\ & =& \left(4.0026-2\times 2.0141\right)u\\ & =& -0.0256 u\end{array}$

We have to calculate the mass difference for the nuclear reaction resulting from 1 gram deuterium.

$\begin{array}{rcl}\Delta m& =& \frac{1g}{2.041g/mol}\times \frac{6.023\times {10}^{23}atoms}{1mol}\times \frac{-0.0256 u}{2 atoms}\times \frac{1\times {10}^{-3}kg}{6.023\times {10}^{23}u}\\ & =& 6.27\times {10}^{-6}kg\end{array}$

Calculating the energy released in fusion reaction

From Einstein’s mass-energy relation, we get:

$\begin{array}{rcl}\Delta E& =& m{c}^2\\ & =& -6.27\times {10}^{-6}kg\times {\left(2.99\times {10}^{8}m/s\right)}^2\\ & =& 5.6\times {10}^{11}J/g\\ & =& 5.6\times {10}^{8}KJ/g\end{array}$

Comparison with energy released from the fission reaction

The energy released in the fission reaction is$7.59\times {10}^7 kJ/g$, while the energy released in the fusion reaction is $5.6\times {10}^8 kJ/g$ .