Physics - Modern Physics Question with Solution | TestHub

Energy released in annihilation $=0.51+0.51=1.02 \mathrm{MeV}$

Initial energy $=1+1=2 \mathrm{MeV}$ .

Therefore, the energy of the two photons is $=1.02+2=3.02 \mathrm{MeV}$ . Hence energy of each photon is $E=1.51 \mathrm{MeV}$ . Now, according to Duane-Hunt law, the wavelength of a photon of energy $E$(in eV) is given by

$\lambda =\frac{\mathit{\text{hc}}}{E\text{in}\text{eV}} \mathrm{\AA }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.51\times {10}^6} \mathrm{\AA }$

$=8.2\times {10}^{-3} \mathrm{\AA }$