Chapter 31, Problem 30P

To determine

To find:The Q-value of the reaction

Answer to Problem 30P

  $4.806\text{MeV}$

Explanation of Solution

Given data:

The reaction is ${}_{92}^{238}\text{U}+\text{n}{\to }_{92}^{239}\text{U}$

The $\mathrm{Q}$ -value for the reaction

  $\begin{array}{c}\mathrm{Q}={\mathrm{m}}_{238}{}_{92}{\text{U}}^2+{\mathrm{m}}_{\text{n}}{\mathrm{c}}^2-{\mathrm{m}}_{239}{\text{u}}^2\\ =\left[238.050788\text{u}+1.008665\text{u}-239.054294\text{u}\right]\left(931.5\frac{\text{MeV}/{\mathrm{c}}^2}{\text{u}}\right){\mathrm{c}}^2\\ =4.806\text{MeV}\end{array}$

Conclusion:

The Q-value of the reaction is $4.806\text{MeV}$