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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 24
Total: 37
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? (1 BeV = 109 eV).
Solution:  
In the process of annihilation, the total energy of electron-positron pair is shared equally by both γ ray photons produced.
Energy of two γ-rays = Energy of electron-positron pair = 10.2 BeV = 10.2 × 109 eV
∴ Energy of each γ-ray photon is
E = 5.1 × 109 eV = 5.1 × 109 × 1.6 × 10–19 J = 5.1 × 1.6 × 10–10 J
But E = hv =
hc
λ
Hence, wavelength associated with γ-ray is
λ =
hc
E
 =
.63×10−34×3×108
5.1×1.6×10−10
 m = 2.44 × 10−16 m
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