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?

Question

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Respuesta :

DeniceSandidge DeniceSandidge · Answer 1 · 2019-08-26T09:50:15+02:00

Answer:

electric potential is 3.31 × [tex]10^{6}[/tex] V

potential energy is 152 MeV

Explanation:

given data

fragment charge Q = 46 protons = 46 × 1.6 × [tex]10^{-19}[/tex] C

to find out

electric potential and potential energy

solution

we know here distance from fragment d = 2 × [tex]10^{-14}[/tex] m

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

so that we can find electric potential = kQ/d

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

electric potential = 3.31 × [tex]10^{6}[/tex] 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 × [tex]10^{6}[/tex]

potential energy = 1.52 × [tex]10^{8}[/tex] eV

so potential energy = 152 MeV