Question

A positron and an electron collide and annihilate to emit two gamma photons of same energy. Calculate the wavelength in pm corresponding to this gamma emission.

(If the answer is X, then write the answer as 2X to the nearest integer).

Answer 1

${}_{+1}^0\text{e}{+}_{-1}^0\text{e}⟶2\gamma$ (photons of same energy)
The mass of two electrons is converted into energy.
The energy produced during emission of two photons
$=2m_e{c}^2=2\times 9.108\times {10}^{-31}\times (3.0\times {10}^8{)}^2$J
$=163.9\times {10}^{-15}$J
$\therefore Energyofonephoton=\frac{163.9\times {10}^{-14}}{2}=8.195\times {10}^{-14}$J
Now, $E=hc/\lambda$
or $8.195\times {10}^{-14}=\frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{\lambda }$
or $\lambda =2.425\times {10}^{-12}m=2.425pm$