Question

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

Answer 1

It is given that the total energy of two y-rays is 10⋅2 BeV or 10⋅2× 10 9 ×1⋅6× 10 −19  J.

The energy for each γ-ray is,

E'= E 2 E'=8⋅16× 10 −19  J

The formula of maximum energy of each γ-ray is,

E'= hc λ

Substitute the values in above expression.

8⋅16× 10 −19 = 6⋅626× 10 −34 ×3× 10 8   λ λ=2⋅436× 10 −16  m

Hence, the value of wavelength associated with each γ-ray is 2⋅436× 10 −16  m.