Question

Calculate the amount of energy released, in kilojoules per gram of uranium, in fission reaction

${}_{\mathbf{92}}{}^{\mathbf{235}}\mathbf{U}\mathbf{+}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathbf{n}\mathbf{\to }{}_{\mathbf{36}}{}^{\mathbf{94}}\mathbf{Kr}\mathbf{+}{}_{\mathbf{56}}{}^{\mathbf{130}}\mathbf{Ba}\mathbf{+}\mathbf{3}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathbf{n}$

Use the atomic masses in Table 19.1. The atomic mass of ${}^{\mathbf{94}}\mathbf{Kr}$is $\mathbf{93}\mathbf{.}\mathbf{919}\mathbf{}\mathbf{u}$and that of ${}^{\mathbf{139}}\mathbf{Ba}$is $\mathbf{138}\mathbf{.}\mathbf{909}\mathbf{}\mathbf{u}$.

Short answer

The energy released in the fission reaction is $7.5\times {10}^7\mathrm{kJ}/\mathrm{g}$.

Step-by-step solution

Calculating the mass difference resulting from the nuclear fission reaction

In a nuclear fission, a heavier nuclide on bombardment with neutrons breaks down into lighter nuclides. The reaction below is an example of a nuclear fission reaction.

${}_{92}{}^{235}\mathrm{U}+{}_0{}^1\mathrm{n}\to {}_{36}{}^{94}\mathrm{Kr}+{}_{56}{}^{139}\mathrm{Ba}+3{}_0{}^1\mathrm{n}$

We have to calculate the mass difference resulting from the nuclear fission reaction:

$\begin{array}{rcl}∆\mathrm{m}& =& \mathrm{m}\left({}_{36}{}^{94}\mathrm{Kr}\right)+\mathrm{m}\left({}_{56}{}^{139}\mathrm{Ba}\right)+3\mathrm{m}\left({}_0{}^1\mathrm{n}\right)-\left({}_{92}{}^{235}\mathrm{U}\right)+\left({}_0{}^1\mathrm{n}\right)\\ & =& \mathrm{m}\left({}_{36}{}^{94}\mathrm{Kr}\right)+\mathrm{m}\left({}_{56}{}^{139}\mathrm{Ba}\right)+2\mathrm{m}\left({}_0{}^1\mathrm{n}\right)-\left({}_{92}{}^{235}\mathrm{U}\right)\\ & =& \left(93.919+138.909+2\times 1.0086-235.043\right)\mathrm{u}\\ & =& -0.2\mathrm{u}\end{array}$

We have to calculate the mass difference corresponding to the decay of 1 g uranium.

$\begin{array}{rcl}∆\mathrm{m}& =& \frac{1\mathrm{g}{}^{235}\mathrm{U}}{235.043\mathrm{g}/\mathrm{mol}}\times \frac{6.023\times {10}^{23}\mathrm{atoms}}{1\mathrm{mol}}\times \frac{-0.2\mathrm{u}}{1\mathrm{atom}}\times \frac{1\times {10}^{-3}\mathrm{kg}}{6.023\times {10}^{23}\mathrm{u}}\\ & =& -8.5\times {10}^7\mathrm{kg}\end{array}$

Calculating the energy released

From Einstein’s mass energy relation, we can find out the energy released in the nuclear fission process.

$\begin{array}{rcl}∆\mathrm{E}& =& {\mathrm{mc}}^2\\ & =& -8.5\times {10}^{-7}\mathrm{kg}\times {\left(2.99\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2\\ & =& -7.59\times {10}^7\mathrm{kg}\end{array}$

The energy released in the fission reaction is$7.59\times {10}^7\mathrm{kJ}/\mathrm{g}$.