A proton of mass `m` moving with a speed `v_(0)` apporoches a stationary proton that is free to move. Assuming impact parameter to be zero., i.e., head-on collision. How close will be incident proton go to other proton ?
A
`(e^2)/(4 pi epsilonm v_(o)^(2))`
B
`( e^2)/(pi epsilon_(0)mv_(o)^(2))`
C
`( e^2)/(mv_(o)^(2))`
D
zero
Text Solution
Verified by Experts
The correct Answer is:
B
Use : LCLM and conservation of energy. i.e., `mv_(0)=2mv-(1)` `(mv_(0)^(2))/(2) = 2xx (1)/(2) m ((v_0)/(2))^2 = ( e^2)/(4 pi epsilon r)`.
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