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A proton of mass m moving with a speed v...

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)/(4piepsilonmv_(0)^(2)`

B

`e^(2)/(piepsilon_(0)mv_(0)^(2)`

C

`e^(2)/(mv_(0)^(2)`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B
use : LCLM and conservation of energy . `Ie:mv_(0)=2mu-(1)`
` (mv_(0)^(2))/2=2xx1/2m(u_(0)/2)=e^(2)/(4piepsilonr)`
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