Answer to Question #313123 in Electricity and Magnetism for melissa

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

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

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

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

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

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

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

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

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