Question 11.24:
In an accelerator experiment on high-energy collisions Of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair Of total energy 10.2 Bev into two \(\gamma\)-rays of equal energy. What is the wavelength associated with each \(\gamma\)- ray? (1 BeV = 109 eV).

Hint: \(E=\frac{hc}{\lambda }\)
Step 1: Find the energy of each \(\gamma\)-ray.

Total energy of two \(\gamma\)-rays, E= 10.2 BeV = 10.2 x 109 eV=10.2 x 109 x 1.6 x 10-19 J.
Hence, the energy of each \(\gamma\)-ray:

E=E2=10.2×1.6×10102=8.16×1010 J

Step 2: Find the wavelength.
Energy is related to wavelength as: 

E=hcλλ=hcE=6.626×1034×3×1088.16×1010=2.436×1016 m

Therefore, the wavelength associated with each \(\gamma\)-ray is 2.436 x 10-16 m.