Question

Assuming an expression for the energy of an electron, derive an expression for the wavelength of spectral lines in a hydrogen atom.

Answer 1

Bohr’s third postulate for the hydrogen atom model says that an atom gives off energy only when an electron moves from a higher energy level to a lower one, and the energy of the light released during this change matches the energy difference between the two electron levels. This radiation emission results in a spectral line.

The energy of the electron in a hydrogen atom, when it is in an orbit with the principal quantum number n, is

`E_n = -(me^4)/(8ε_0^2h^2n^2)`

where m = mass of the electron, e = electronic charge, h = Planck’s constant and ε₀ = permittivity of free space.

Let E_m be the energy of the electron in a hydrogen atom while it is in an orbit with the principal quantum number m and E_n be its energy in an orbit with the principal quantum number n, where n < m. Then

`E_m = -(me^4)/(8ε_0^2h^2m^2)` and `E_n = -(me^4)/(8ε_0^2h^2n^2)`

Therefore, the energy radiated when the electron jumps from the higher energy state to the lower energy state is

`E_m - E_n = -(me^4)/(8ε_0^2h^2m^2) - (-(me^4)/(8ε_0^2h^2n^2))`

= `(me^4)/(8ε_0^2h^2) (1/n^2 - 1/m^2)`

This energy is emitted in the form of a quantum of radiation (photon) with energy hν, where ν is the frequency of the radiation.

∴ E_m − E_n = hν

∴ `ν = (E_m - E_n)/h = (me^4)/(8ε_0^2h^3) (1/n^2 - 1/m^2)`

The wavelength of the radiation is λ = `c/ν`, where c is the speed of radiation in free space.

The wave number, `barν = 1/λ = ν/c`

∴ `barν = 1/λ = (me^4)/(8ε_0^2h^3c) (1/n^2 - 1/m^2) = R(1/n^2 - 1/m^2)`

where R `(= (me^4)/(8ε_0^2h^3c))` is a constant called the Rydberg constant.

This expression gives the wave number of the radiation emitted and hence that of a line in the hydrogen spectrum.

For the Lyman series, n = 1, m = 2, 3, 4, ... ∞.

∴ `1/λ_L = R (1/1^2 - 1/m^2)` and for the shortest wavelength line in this series, `1/λ_(Ls) = R(1/1^2)` as m = ∞.

For the Balmer series, n = 2, m = 3, 4, 5, ... ∞.

∴ `1/λ_B = R(1/4 - 1/m^2)` and for the shortest wavelength line in this series, `1/λ_(Bs) = R(1/4)` as m = ∞.