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Question: One MeV positron encounters 1 MeV electron travelling in opposite direction. What is the wavelength ...

One MeV positron encounters 1 MeV electron travelling in opposite direction. What is the wavelength of photons produced? (Given rest mass energy of electron) or positron = 0.512MeV0.512MeV and h=6.62×10−34h = 6.62 \times 10^{- 34}Js

A

8.2×10−11m8.2 \times 10^{- 11}m

B

8.2×10−13m8.2 \times 10^{- 13}m

C

8.2×10−12m8.2 \times 10^{- 12}m

D

8.2×10−9m8.2 \times 10^{- 9}m

Answer

8.2×10−13m8.2 \times 10^{- 13}m

Explanation

Solution

: −10e++10e→2γ{}_{- 1}^{0}e +_{+ 1}^{0}e \rightarrow 2\gamma

Energy of each photon

=2(1+0.512)2=1.512MeV=1.512×1.6×10−13J= \frac{2(1 + 0.512)}{2} = 1.512MeV = 1.512 \times 1.6 \times 10^{- 13}J

As E=hcλE = \frac{hc}{\lambda}

∴λ=hcE=6.62×10−34×3×1081.512×1.6×10−13=8.2×10−3m\therefore\lambda = \frac{hc}{E} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{1.512 \times 1.6 \times 10^{- 13}} = 8.2 \times 10^{- 3}m