Question

In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton. How many revolutions per second does the electron make?

Short answer

Revolutions made by the electron per second is $6.57\times {10}^{15}$revolutions/s.

Step-by-step solution

Step 1. Given information

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

Step 2.Formulas used

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

Therefore, $F_e=F_{cp}—1$

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

k is Coulomb constant.

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

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

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

Step 3.Calculation

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

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

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

$\to w=\sqrt{k·\frac{\left|q1·q2\right|}{m·r^3}}—4$

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

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

$\frac{w}{2\mathrm{\pi }}=6.57·{10}^{15}$

Step 4.Final answer

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

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