Question

Calculate the electrostatic force of attraction between a proton and an electron in a hydrogen atom. The radius of the electron orbit is 0.05 nm and charge on the electron is $1.6\times {10}^{-9}C.$

Answer 1

From Coulomb's inverse square law: $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
The magnitude of charge on electron and proton is the same $=1.6\times {10}^{-19}C$
Therefore, electrostatic force is given by
$F=\frac{9\times {10}^9\times 1.6\times {10}^{-19}\times 1.6\times {10}^{-19}}{(0.05\times {10}^{-9}{)}^2}$
$=\frac{9\times 1.6\times 1.6\times {10}^{-29}}{(5\times {10}^{-11}{)}^2}$
$=\frac{9\times 1.6\times 1.6\times {10}^{-29}}{25\times {10}^{-22}}$
$=0.9216\times {10}^{-29}\times {10}^{+22}$
$=0.9216\times {10}^{-7}N$
$=9.216\times {10}^{-8}N$