Question

In a hydrogen atom (i.e., one stationary proton and one orbiting electron), the angular velocity of the electron is 10 x 10^6 radians per second. What is the radius of the electron orbit? Assume uniform circular motion.

Answer 1

Assuming the centripetal force is provided by the Coulomb attraction between the electron and the proton we have that,

$F_c = k \frac{e^2}{R^2}$

Here,

k = Coulomb's Constant

R = Distance

e = Electron charge

And by the centripetal force,

$F_c = m_e R\omega^2$

$m_e$ = Mass of electron

R = Radius

$\omega$ = Angular velocity

Equation both expression,

$k \frac{e^2}{R^2} = m_e R\omega^2$

Replacing,

$R^3 = \frac{(9*10^9)(1.6*10^{-19})^2}{(9.11*10^{-31})(10^{6})^2}$

$R^3 = 2.52908*10^{-10} m$

$R = 6.32394*10^{-4}m$

Therefore the radius of the electron orbit is $6.32394*10^{-4}m$