Question

In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)

Answer 1

in the process of annihilation of an electron, the total energy of electron - positron pair is shared equally by both $\gamma$- ray photons produced.

so, energy of two $\gamma$ - rays = energy of electron - positron pair = 10.2 BeV = 10.2 × 10^9 eV

so, energy of each $\gamma$ - ray photon, E = 10.2 × 10^9/2 eV = 5.1 × 10^9 eV

we know, 1eV = 1.6 × 10^-19 J

so, E = 5.1 × 1.6 × 10^-10 J

now, wavelength associated with $\gamma$-ray , λ = hc/E

= (6.63 × 10^-34 × 3 × 10^8)/(5.1 × 1.6 × 10^-10)

= 2.44 × 10^-16 m

hence, wavelength associated with each $\gamma$-ray = 2.44 × 10^-16 m