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A photon of energy 1.12 MeV splits into ...

A photon of energy `1.12 MeV` splits into electron positron pair. The velocity of electron is (Neglect relativistic correction)

A

`3xx10^(8) ms^(-1)`

B

`1.33xx10^(8) ms^(-1)`

C

`6xx10^(8) ms^(-1)`

D

`9xx10^(8) ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(gamma)=((1)/(2)mv^(2))2+2E_(0)`
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