Question

Find the magnitude of the electrostatic force of attraction between an iron nucleus, Fe and its innermost orbiting electron if the distance of separation between them is 1.50 x 10^-12 m.  Determine also the orbital speed of this electron.

Answer 1

Force

$F=\frac{kq_1q_2}{r^2}$

$F=\frac{kze^2}{r^2}$

$F=\frac{9\times10^9\times26\times1.6^2\times10^{-38}}{1.50^2\times10^{-24}}$

$F=2.67\times10^{-3}N$

$\frac{mv^2}{r}=\frac{kze^2}{r^2}$

$v=\sqrt\frac{{kze^2}}{rm}$

$v=\sqrt\frac{{9\times10^9\times26\times(1.6\times10^{-19})^2}}{1.50\times10^{-12}\times9.1\times10^{-31}}$

$v=6.62\times10^{7} m/sec$