Question

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?

Answer 1

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=momega^2r=m4pi^2n^2r`

where radius of the orbit `" "r=4.76xx10^-10m`

mass of an electron `" "m=9.1xx10^-31kg`

`F=1.02xx10^-19N`

So

`m4pi^2n^2r=F`

`=>n^2=F/(4pi^2xxmxxr)`

`=>n=sqrt(F/(4pi^2xxmxxr))`

`=sqrt((1.02xx10^-19N)/(4pi^2xx9.1xx10^-31kgxx4.76xx10^-10m))`

`=3.74xx10^8Hz`