Question

9.1.Find the force of attraction betweenthenucleus of a hydrogen

atom and an electron. The radius of the hydrogen atom isO.5x10-8cm

and the charge of the nucleus is equal in magnitude and opposite in

sign to that of the electron.

Answer 1

Solution

Charge of electron

Q$=1.6\times10^{-19}C$

Radius

R=O.5x10^-8 cm

Then electrostatic force between electron and nucleus

$F=\frac{KQQ}{R^2}\\=\frac{9\times10^9\times1.6\times10^{-19}\times 1.6\times10^{-19} }{( 0.5\times10^{-10}) ^2}\\=92.16\times10^{-9}N.$