Chapter 30.3, Problem 38PP

To determine

To calculate:Energy of the third gamma ray.

Answer to Problem 38PP

The energy of third gamma ray is $1458\text{keV}$ .

Explanation of Solution

Given:

For a positron and electron,

Energy of first gamma $\left(E_{\gamma 1}\right)$ : $357\text{keV}$

Energy of first gamma $\left(E_{\gamma 2}\right)$ : $225\text{keV}$

Calculation:

The energy of positron and electron is given by,

  $E=2\times \left(1.02\text{MeV}\right)\text{=}\text{2}\text{.04}\text{MeV}$

Substitute all the given values in the formula,

  $E_{\gamma 3}+E_{\gamma 2}+E_{\gamma 1}=2040\text{keV}$

  $E_{\gamma 3}=2040-225-357=1458\text{keV}$

  $E_{\gamma 3}=1458\text{keV}$

Conclusion:

Hence, the energy of third gamma ray is $1458\text{keV}$ .