Protons are accelerated through a potential difference of 6.0 MV (mega volts) and?

Question

Protons are accelerated through a potential difference of 6.0 MV (mega volts) and then make head-on collisions with atomic nuclei with a charge of +82 elementary charges. What is the closest distance of approach between the protons and the nuclei? Assume the nuclei are stationary.

Answer: 2.0 x 10^-14 m

Answer 1

When a particle of charge "q" is accelerated through P.D. "V". it acquires Kinetic Energy:-

    K.E.   =   q x V

Here, K.E. of a proton will be    1.6 x10^-19 x 6 x 10^6

                                                       =   9.6 x 10 -13  Joules.

The proton approaches the nucleus, and stops at a distance "r" away from it due to electrostatic repulsion.

At this point, the K.E. the proton had will have been converted to Electrical Potential Energy

Electrical potential at a distance "r" from charge "q"

        =       q / 4 * pi * eo * r       where eo   =   permittivity of free space

Electrical Potential Energy   when charge Q is at a distance "r form charge "Q":-

         =     q * Q / 4 * pi * eo * r

The Kinetic Energy of the proton will be equal to this.

So   (9.6 *10^-13 )    =     (1.6 *10^-19 ) * 82 * (1.6 * 10^-19) / 4 * pi * (8.85 * 10^-12) * r

so  r   =    (1.6 * 10^-19) * 82 * (1.6 * 10^-19) / 4 * pi * (8.85 * 10^-12) * (9.6 * 10^-13)

     r   =   1.996 * 10^-14 m

    say   r  =  2.0 * 10^-14 m.

Answer 2

i dont get it