Question

In nuclear fission, a nucleus splits roughly in half. What is the potential 5.00×10 –12 m from a fragment that has 42 protons in it?

Answer 1

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

$q=42e=42\times1.6\times10^{-19}C$

$V=\frac{9\times10^9\times42\times1.6\times10^{-19}}{5\times10^{-12}}=12096V$