Question

Show that the angular speed of an electron in the nth Bohr orbit is w = `(πme^4)/(2ε_0^2h^3n^3)` and the corresponding frequency of the revolution of the electron is f = `(me^4)/(4ε_0^2h^3n^3)`.

Answer 1

The radius of the nth Bohr orbit is

`r = (ε_0h^2n^2)/(πmZe^2)`  ...(1)

and the linear speed of an electron in this orbit is

`v = (Ze^2)/(2ε_0nh)`  ...(2)

where ε₀ ≡ permittivity of free space, h ≡ Planck’s constant, n ≡ principal quantum number, m ≡ electron mass, e ≡ electronic charge and Z ≡ atomic number of the atom. v Since angular speed ω = `v/r`, from Eqs. (1) and (2), we get

ω = `v/r`

= `(Ze^2)/(2ε_0nh). (πmZe^2)/(ε_0h^2n^2)`

= `(πmZ^2e^4)/(2ε_0^2h^3n^3)`  ...(3)

which gives the required expression for the angular speed of an electron in the nth Bohr orbit. From Eq. (3), the frequency of revolution of the electron,

`f = ω/(2π)`

= `1/(2π) xx (πmZ^2e^4)/(2ε_0^2h^3n^3)`

= `(mZ^2e^4)/(4ε_0^2h^3n^3)`

as required.