Question

A $1\text{MeV}$ positron and a $1\text{MeV}$ electron meet, each moving in opposite directions. They annihilate each other by emitting two photons. If the rest mass energy of an electron is $0.51\text{MeV}$, then the wavelength of each photon is

• A. $5.1\times {10}^{-3}\mathring{A}$
• B. $10.2\times {10}^{-3}\mathring{A}$
• C. $8.2\times {10}^{-3}\mathring{A}$
• D. $6.2\times {10}^{-3}\mathring{A}$

Answer 1

The correct option is C $8.2\times {10}^{-3}\mathring{A}$

The total $K.E$ of electron and position $\left(E_{KE}\right)=2\text{MeV}$

The total rest mass energy of electron and position
$E_{mass}=0.51+0.51=1.02\text{MeV}$

Let $E_p$ be the energy of a photon,

Then, By conservation of energy,

$2E_p=E_{K.E}+E_{mass}$

$\Rightarrow 2E_p=2+1.02=3.02\text{MeV}$

$\Rightarrow E_p=1.51\text{MeV}$

The wavelength of one photon is

$E_p=h\gamma =\frac{hc}{\lambda }$

$\Rightarrow \lambda =\frac{hc}{E}=\frac{12400\text{eV}\mathring{A}}{1.51\times {10}^6\text{eV}}$

$\Rightarrow \lambda =8.2\times {10}^{-3}\mathring{A}$

Hence, option $\left(C\right)$ is correct.

Why this question? To know and understand about the pair annihilation of particles like electrons and positrons.