Chapter 32, Problem 26P

To determine

The wavelength of photon produced during head on collision.

Answer to Problem 26P

  $\mathrm{\lambda }=0.0133\times {10}^{-10}\text{ m}$

Explanation of Solution

Given:

  ${\mathrm{e}}^{+}+{\mathrm{e}}^{-}\to 2\mathrm{\gamma }$

Rest energy of electron and positron, ${\mathrm{R}}_{\mathrm{E}}=0.511\text{ MeV}$

Kinetic energy of electron and positron, $\mathrm{K}.\mathrm{E}=420\text{ keV}$

Formula used:

Total energy is, $\mathrm{E}={\mathrm{R}}_{\mathrm{E}}+\mathrm{K}.\mathrm{E}$

Where, $\mathrm{K}.\mathrm{E}$ = kinetic energy

  ${\mathrm{R}}_{\mathrm{E}}$ = rest energy

  $\mathrm{E}$ = total energy

Calculation:

According to the principle of conservation of energy, the sum of rest and kinetic energy of electron and positron is equal to total energy.

  $\begin{array}{l}\mathrm{E}={\mathrm{R}}_{\mathrm{E}}+\mathrm{K}.\mathrm{E}\\ \mathrm{E}=\left(0.511\times {10}^6+420\times {10}^3\right)\text{ eV}\\ \mathrm{E}=931\text{ keV}\end{array}$

The wavelength is calculated as,

  $\mathrm{\lambda }=\frac{\mathrm{h}\mathrm{c}}{\mathrm{E}}$

Where, $\mathrm{\lambda }$ = wavelength

  $\mathrm{E}$ = total energy

  $\mathrm{c}$ = speed of light

  $\mathrm{h}$ = Plank's constant

  $\begin{array}{l}\mathrm{\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{\left(931\times {10}^3\times 1.6\times {10}^{-19}\right)}\\ \mathrm{\lambda }=0.0133\times {10}^{-10}\text{ m}\end{array}$