In nuclear fission, a nucleus splits roughly in half. (a) what is the potential 2.00 × 10−14 m from a fragment that has 46 protons in it? (b) what is the potential energy in mev of a similarly charged fragment at this distance?

electric potential is 3.31 × V

potential energy is 152 MeV

Explanation:

given data

fragment charge  Q = 46 protons = 46 × 1.6 × C

to find out

electric potential  and potential energy

solution

we know here distance from fragment d = 2 × m

and constant for electric force k that is 9 × N-m²/C²

so that we can find electric potential = kQ/d

electric potential = 9 × / ( 2 × )

electric potential = 3.31 × V

and

we know relation between electric potential and potential

that is  V = U/q

so U will be = qV

now put all value

we get potential energy U

potential energy = 46 × 3.31 ×

potential energy = 1.52 × eV

so potential energy = 152 MeV