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 ?

Options

A. `(e^(3))/(pi epsilon_(0) m^(2) v_(0))`
B. `(e^(3))/(pi epsilon_(0) m v_(0))`
C. `(e^(2))/(pi epsilon_(0) m v_(0)^(2))`
D. None of these

Answer

C

Explanation

The protons move loward each other till their relative velocity becomes equal to zero. At the closest distance of approach, both the proton will be moving with the same velocity .
As coulembian repulsive force is internal for the sysytem of photon, we can apply the law of conservation of momentum ,
`:. m v_(0) = 2 m v`
Change in `KE = (1)/(2) m v_(0)^(2) - 2 xx (1)/(2) m ((v_(0))/(2))^(2)`
This change in energy is equal to the electrical potential energy
`(m v_(0)^(2))/(2) = m ((v_(0))/(2))^(2) = (e^(2))/(4 pi epsilon_(0) r)`
`:. r = (e^(2))/(pi epsilon_(0)m v_(0)^(2))`