Answer to Question #207902 in Atomic and Nuclear Physics for Tuyen Vu

Question

What is the distance between two protons so that their Coulomb potential is barrier potential 200 keV 

Answer 1

Gives

V=200Kev

$V=\frac{Kz_1z_2e^2}{r}$

For coulamb potential (Alpha partical)

$z_1=z_2=2$

$r=\frac{kz_1z_2e^2}{V}$

Put value

$$r=\frac{9\times10^9\times2\times 2\times(1.6\times10^{-19})^2}{200\times1.6\times10^{-16}}=0.288\times10^{-10}$$