Question

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?

Answer 1

Answer:

electric potential is 3.31 × $10^{6}$ V

potential energy is 152 MeV

Explanation:

given data

fragment charge  Q = 46 protons = 46 × 1.6 × $10^{-19}$ C

to find out

electric potential  and potential energy

solution

we know here distance from fragment d = 2 × $10^{-14}$ m

and constant for electric force k that is 9 × $10^{9}$ N-m²/C²

so that we can find electric potential = kQ/d

electric potential = 9 × $10^{9}[/tex ×46 × 1.6 × [tex]10^{-19}$ / ( 2 × $10^{-14}$ )

electric potential = 3.31 × $10^{6}$ 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 × $10^{6}$

potential energy = 1.52 × $10^{8}$ eV

so potential energy = 152 MeV