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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 24 of 37
Marks: +1, -0
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 = 10910^{9} 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 × 10910^{9} eV
∴ Energy of each γ-ray photon is
E = 5.1 × 10910^{9} eV = 5.1 × 10910^{9} × 1.6 × 10−1910^{-19} J = 5.1 × 1.6 × 10−1010^{-10} J
But E = hv = hcλ\frac{hc}{\lambda}
Hence, wavelength associated with γ-ray is
λ = hcE\frac{hc}{E} = .63×10−34×3×1085.1×1.6×10−10\frac{.63 \times 10^{-34} \times 3 \times 10^{8}}{5.1 \times 1.6 \times 10^{-10}} m = 2.44 × 10−1610^{-16} m
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