Chapter 31, Problem 19P

To determine

The rest mass of ${}_{92}^{235}\text{U}$ .

Answer to Problem 19P

The rest mass of ${}_{92}^{235}\text{U}$ is $\frac{1}{1100}$ .

Explanation of Solution

Given:

The energy produced in the reactor is of 200 MeV

Formula used:

The expression is given by,

  $\mathrm{n}=\frac{\left(\frac{\mathrm{Q}}{931.5}\right)}{\mathrm{m}}$

Here

• m is 235.043923, which is mass of ${}_{92}^{235}\text{U}$

Calculation:

Substitute the values in the above equation.

  $\begin{array}{c}\mathrm{n}=\frac{\left(\frac{\mathrm{Q}}{931.5}\right)}{\mathrm{m}}\\ =\frac{\left(\frac{200}{931.5}\right)}{235.043923}\\ =\frac{1}{1100}\end{array}$

Thus, the rest mass of ${}_{92}^{235}\text{U}$ is $\frac{1}{1100}$ .