Question

What is the origin of spectral lines? Obtain an expression for the wave number of a line in hydrogen spectrum.

Answer 1

Bohr's third postulate for the hydrogen atom model states that an atom can only emit energy when an electron jumps from a higher energy state to a lower energy state and that the energy difference between the two states of the electron equals the energy of the quantum of electromagnetic radiation emitted during this process. A spectral line is produced as a result of this radiation release. When a hydrogen atom's electron is in an orbit with the principal quantum number n, its energy equals

`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 represent the electron's energy in an orbit with principle quantum number m a hydrogen atom, and let E_n represent the electron's energy in an orbit with principal quantum number n, n < m. Next

`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 h, where hv 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, `overline(ν) = 1/λ = ν/c`.

∴ `overline(ν) = 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.