# In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.30 x 10^6 m/s. How would you find the force acting on the electron as it revolves in a circular orbit of radius 5.30 x10^ -11 m?

Nov 3, 2015

You would find it like this:

#### Explanation:

I will apply classical physics to this problem.

The centripetal force is given by:

$F = \frac{m {v}^{2}}{r}$

I'll use the rest mass of the electron:

$m = 9.1 \times {10}^{- 31} \text{kg}$

$\therefore F = \frac{9.1 \times {10}^{- 31} \times {\left(2.3 \times {10}^{6}\right)}^{2}}{5.3 \times {10}^{- 11}}$

$F = 9.08 \times {10}^{- 8} \text{N}$

This is not a good question as it is not compatible with the Bohr model of the atom. An electron in a circular orbit like this would emit radiation and, by losing energy, spiral into the nucleus.

To overcome this problem Bohr proposed the concept of "energy levels".

I have given an answer based on classical physics.