A proton and an antiproton come close to each other in vacuum such that the distance between them is 10   cm . Consider the potential energy to be zero at infinity. The velocity at this distance will be:

Step-by-step solution

Applying mechanical energy conservation K i + U i = K f + U f 0 + 0 = 1 2 m v 2 + 1 2 m v 2 + k q 1 q 2 r q 1 = q 2 = e ,    q 1 = + e q 2 = - e and r = 0 . 1   m ∴ m v 2 = k e 2 r ∴ v 2 = k e 2 m r = 9 × 10 9 × 1 . 6 × 10 - 19 2 1 . 67 × 10 - 27 ( 0 . 1 ) v ≃ 1 . 17   m / s