Answer to Question #157576 in Atomic and Nuclear Physics for syeda sana

Question

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?

Answer 1

The electric potential due to protons at distance r is

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

$=\frac{(9\times 10^9 N.m^2/C^2)(53\times1.60\times10^{-19}C)}{4.50×10^{–14}}$

$=1.696 \times 10^6 V$

The electrostatic potential energy is

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

$=\frac{(9\times 10^9 N.m^2/C^2)(53\times1.60\times10^{-19}C)^2}{4.50×10^{–14}m}$

$=1.438 \times 10^{-11} J$