Question #157576

A nuclear fission reaction is defined to be a process in which a large nucleus divides into two

roughly equal parts.

(i) What is the potential at a distance of 4.50×10–14 m from one of the parts that has 53 protons in it?

(ii) What is the potential energy in eV of a similarly charged part at this distance?


1
Expert's answer
2021-01-22T11:27:12-0500

The electric potential due to protons at distance r is

V=kQrV=\frac{kQ}{r}

=(9×109N.m2/C2)(53×1.60×1019C)4.50×1014=\frac{(9\times 10^9 N.m^2/C^2)(53\times1.60\times10^{-19}C)}{4.50×10^{–14}}

=1.696×106V=1.696 \times 10^6 V

The electrostatic potential energy is

U=kQ2rU=\frac{kQ^2}{r}

=(9×109N.m2/C2)(53×1.60×1019C)24.50×1014m=\frac{(9\times 10^9 N.m^2/C^2)(53\times1.60\times10^{-19}C)^2}{4.50×10^{–14}m}

=1.438×1011J=1.438 \times 10^{-11} J


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