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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)
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Solution
Total energy of two γ-rays:
E = 10. 2 BeV
= 10.2 × 109 eV
= 10.2 × 109 × 1.6 × 10−10 J
Hence, the energy of each γ-ray:
`"E'" = "E"/2`
= `(10.2 xx 1.6 xx 10^(-10))/2`
= 8.16 × 10−10 J
Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
Energy is related to wavelength as:
`"E'" = "hc"/lambda`
∴ `lambda= "hc"/"E'"`
= `(6.626 xx 10^(-34) xx 3 xx 10^8)/(8.16 xx 10^(-10))`
= 2.436 × 10−16 m
Therefore, the wavelength associated with each γ-ray is 2.436 × 10−16 m.
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