An electron and a positron undergo pair annihilation (Eq. 44-5). If they had approximately zero kinetic energy before the annihilation, what is the wavelength of each$\mathit{\gamma }$produced by the annihilation?

Annihilation.

When a particle joins its antiparticle, the two of them can annihilate each other. The particle and antiparticle disappear, and their combined energies reappear in some other forms, which are equally shared by the two protons.

If the electron and positron are stationary when annihilating, then their totalenergy is their total mass energy.

The Annihilation of electron and proton can be given as

$e^{-}+e^{+}=\gamma +\gamma$

We know that the rest energy of the electron,

$E=m_e{c}^2=0.511 \mathrm{MeV}$

Since the mass of both particles is the same, they have the same rest or mass energy.

The total rest energy of the electron-positron pair is

$2E=2\left(0.511 \mathrm{MeV}\right)$

Therefore, the energy of each $\gamma$ produced

$\begin{array}{l}2E_{\gamma }=2E=2\left(0.511 \mathrm{MeV}\right)\\ E_{\gamma }=0.511 \mathrm{MeV}\end{array}$

We know that wavelength can be given as

$\lambda =\frac{hc}{E}$

Where E is energy, c is the velocity of light, h is the plank constant, And the standard value of hc = 1240 eV .nm

Now substituting all the values in the formula

$\begin{array}{c}\lambda =\frac{hc}{E_{\gamma }}\\ =\frac{1240 \mathrm{eV}. \mathrm{nm}}{0.511\times {10}^{-6} \mathrm{eV}}\\ =2.43 \times {10}^{-3} \mathrm{nm}\\ =2.43 \mathrm{pm}\end{array}$

Therefore, the wavelength of each$\gamma$produced is 2.43 pm.