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kherson [118]
3 years ago
5

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 proto

ns in it? (b) What is the potential energy in MeV of a similarly charged fragment at this distance?
Physics
1 answer:
Valentin [98]3 years ago
4 0

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

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