Question

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately 2.15 106 m/s?

Answer 1

This is an incomplete question, here is a complete question.

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately $2.15\times 10^6m/s$?

Find the force acting on the electron as it revolves in a circular orbit of radius $5.34\times 10^{-11}m$.

Answer : The force acting on electron is, $7.9\times 10^{-8}N$

Explanation :

Formula used :

$F=\frac{mv^2}{r}$

where,

F = force acting on electrons

m = mass of electrons = $9.1\times 10^{-31}kg$

v = speed of electron = $2.15\times 10^6m/s$

r = radius of circular orbit = $5.34\times 10^{-11}m$

Now put all the given values in the above formula, we get:

$F=\frac{(9.1\times 10^{-31}kg)\times (2.15\times 10^6m/s)^2}{5.34\times 10^{-11}m}$

$F=7.9\times 10^{-8}N$

Thus, the force acting on electron is, $7.9\times 10^{-8}N$