91影视

A hydrogen atom and the radius of the electron's orbit is 0.053 nm.

As the electron and proton are charged an electric force acts. And electron moves in circle around proton so, centripetal force acts.

Therefore, $F_e=F_{cp}鈥�1$

Electric field$\left(F_e=k路\frac{\left|q_1路q_2\right|}{r^2}\right)鈥�2$,$\left(k=9路{10}^9\frac{N{m}^2}{C^2}\right)$

k is Coulomb constant.

Centripetal force$\left(F_{cp}=m路r路w^2\right)鈥�3$,

$\left(m=9.11路{10}^{-31}kg\right)$m is mass of the electron.

Charge of electron and proton are equal, i.e.$\left|q1\right|=\left|q2\right|=1.602路{10}^{-19}C$.

One revolution will be 2$\mathrm{蟺}$,so the angular velocity will be divided by 2$\mathrm{蟺}$.

So, including formula 2 and 3 in eq1 we get,

$k路\frac{\left|q1路q2\right|}{r^2}=m路r路w^2$

$鈫�w=\sqrt{k路\frac{\left|q1路q2\right|}{m路r^3}}鈥�4$

Now dividing eq4 by 2$\mathrm{蟺}$we get,

$\frac{w}{2\mathrm{蟺}}=\frac{1}{2\mathrm{蟺}}\sqrt{k路\frac{\left|q1路q2\right|}{m路r^3}}=\frac{1}{2\mathrm{蟺}}\sqrt{9路{10}^9路\frac{{\left(1.602路{10}^{-19}\right)}^2}{9.11路{10}^{-31}路{\left(0.053路{10}^{-9}\right)}^3}}$

$\frac{w}{2\mathrm{蟺}}=6.57路{10}^{15}$

$\frac{w}{2\mathrm{蟺}}=6.57路{10}^{15}$revolutions/s.

i.e. revolutions made per second is 6.57$路$10¹⁵.