Chapter 31, Problem 20P

(a)

To determine

The number of neutrons produced.

Answer to Problem 20P

The number of neutrons produced is 5

Explanation of Solution

Given:

Consider the reaction.

  ${}_{92}^{235}\text{U}+\text{n}{\to }_{51}^{133}\text{Sb}{+}_{41}^{98}\text{Nb}+?\text{n}$

Calculation:

Calculate the mass number from both the sides.

  $\begin{array}{l}235+1=133+98+?\\ ?=5\end{array}$

Thus, the number of neutrons produced is 5.

(b)

To determine

The released energy.

Answer to Problem 20P

The released energy.is 171.1 MeV

Explanation of Solution

Given:

The atomic mass of Sb and Nb isotopes are 132.915250u and 97.910328u.

Formula used:

The reaction is,

  ${}_{92}^{235}\text{U}+\text{n}{\to }_{51}^{133}\text{Sb}{+}_{41}^{98}\text{Nb}+5\text{n}$

Thus,

  $\mathrm{Q}={\mathrm{m}}_{{}_{92}^{235}\text{U}}{\mathrm{c}}^2+{\mathrm{m}}_{\text{n}}{\mathrm{c}}^2-{\mathrm{m}}_{{}_{51}^{133}\text{Sb}}{\mathrm{c}}^2-{\mathrm{m}}_{{}_{41}^{98}\text{Nb}}{\mathrm{c}}^2-5{\text{m}}_{\mathrm{n}}{\mathrm{c}}^2$

Calculation:

Substitute the values in the above equation.

  $\begin{array}{c}\mathrm{Q}={\mathrm{m}}_{{}_{92}^{235}\text{U}}{\mathrm{c}}^2+{\mathrm{m}}_{\text{n}}{\mathrm{c}}^2-{\mathrm{m}}_{{}_{51}^{133}\text{Sb}}{\mathrm{c}}^2-{\mathrm{m}}_{{}_{41}^{98}\text{Nb}}{\mathrm{c}}^2-5{\text{m}}_{\mathrm{n}}{\mathrm{c}}^2\\ =235.04392\mathrm{u}+1.00865\mathrm{u}-132.9152\mathrm{u}-97.9103\mathrm{u}-5\left(1.008665\right)\left(93.5\right)\\ =171.1\text{ MeV}\end{array}$

Thus, the released energy.is 171.1 MeV.