# In the Bohr model of the hydrogen atom, an electron (mass = 9.10 10-31 kg) orbits a proton at a distance of 4.76x10-10 m. The proton pulls on the electron with an electric force of 1.02x10-9 N. How many revolutions per second does the electron make?

Dec 10, 2016

We know that the force of attraction $F$of proton on electron in the hydrogen atom provides the required centripetal force required for the motion of electron in circular orbit.

If $n$ be the number of revolutions per sec made by the electron around the nucleus then centripetal force is given by

$F = m {\omega}^{2} r = m 4 {\pi}^{2} {n}^{2} r$

where radius of the orbit $\text{ } r = 4.76 \times {10}^{-} 10 m$

mass of an electron $\text{ } m = 9.1 \times {10}^{-} 31 k g$

$F = 1.02 \times {10}^{-} 19 N$

So

$m 4 {\pi}^{2} {n}^{2} r = F$

$\implies {n}^{2} = \frac{F}{4 {\pi}^{2} \times m \times r}$

$\implies n = \sqrt{\frac{F}{4 {\pi}^{2} \times m \times r}}$

$= \sqrt{\frac{1.02 \times {10}^{-} 19 N}{4 {\pi}^{2} \times 9.1 \times {10}^{-} 31 k g \times 4.76 \times {10}^{-} 10 m}}$

$= 3.74 \times {10}^{8} H z$