Chapter 27, Problem 37P

To determine

The energy and wavelength of the photon that produces an electron and a positron.

Answer to Problem 37P

Solution:

The photon energy is $1.6\text{MeV}$.

Wavelength of the photon is $7.79\times {10}^{-13}\text{m}$.

Explanation of Solution

Given:

Kinetic energy of the particles:$285\text{keV}$

Formula Used:

$hf=\frac{hc}{\lambda }=2m_0{c}^2$

Calculation:

The energy of photon can be calculate as

$\begin{array}{c}{E}_{ph}=2\left(\text{KE+}{m}_0{c}^2\right)\\ =2(0.285\text{MeV}+0.511\text{MeV})\\ =1.592\text{MeV}\approx 1.6\text{MeV}\end{array}$

The wavelength of the photon can be given as

$\begin{array}{l}E=\frac{hc}{\lambda }\\ \text{or,}\lambda \text{=}\frac{hc}{E}=\frac{4.1357\times {10}^{-15}\text{eV}\text{s}\times 3\times {10}^8\text{m/s}}{\text{1}\text{.6}\times {\text{10}}^6\text{eV}}=7.79\times {10}^{-13}\text{m}\end{array}$