Chapter 31, Problem 17P

To determine

The energy released in the fission reaction.

Answer to Problem 17P

The energy released in the fission reaction is $173.3\text{ MeV}$ .

Explanation of Solution

Given:

The mass of ${}_{56}^{141}{\text{Ba and }}_{36}^{92}\text{Kr}$ is $140.914411\text{ u and 91}\text{.926156 u}$ .

Formula used:

Use Appendix B,

Calculate the required energy.

  $\text{Q}={\mathrm{m}}_{\mathrm{n}}{\mathrm{c}}^2+{\mathrm{m}}_{{}_{92}^{235}}{}_{\text{U}}{\mathrm{c}}^2-{\mathrm{m}}_{{}_{56}^{141}\text{Ba}}+{\mathrm{m}}_{{}_{36}^{92}\text{Kr}}{\mathrm{c}}^2-3{\mathrm{m}}_{\mathrm{n}}{\mathrm{c}}^2$

Calculation:

Substitute the values in the above equation.

  $\begin{array}{c}\text{Q}={\mathrm{m}}_{\mathrm{n}}{\mathrm{c}}^2+{\mathrm{m}}_{{}_{92}^{235}}{}_{\text{U}}{\mathrm{c}}^2-{\mathrm{m}}_{{}_{56}^{141}\text{Ba}}+{\mathrm{m}}_{{}_{36}^{92}\text{Kr}}{\mathrm{c}}^2-3{\mathrm{m}}_{\mathrm{n}}{\mathrm{c}}^2\\ =\left[1.008665\mathrm{u}+235.0439\mathrm{u}-140.9141\mathrm{u}-91.9356156\mathrm{u}-3\left(1.008665\mathrm{u}\right)\left(931.5{\mathrm{c}}^2\right)\right]\\ =173.3\text{ MeV}\end{array}$

Thus, the energy released in the fission reaction is $173.3\text{ MeV}$ .