Question

Starting from the formula for the energy of an electron in the nth orbit of the hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.

Answer 1

An atom emits energy only when an electron jumps from a higher energy state to a lower energy state, according to Bohr's third postulate for the model of the hydrogen atom, and the energy of the quantum of electromagnetic radiation emitted in this process is equal to the energy difference between the two states of the electron.

A spectral line is formed as a result of this radiation emission. When an electron in a hydrogen atom is in an orbit with the principal quantum number n, its energy is

`"E"_"n" = - "me"^4/(8ε_0^2"h"^2"n"^2)`

where, m = mass of 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 when it is in an orbit with the principal quantum number m and E_n, its energy in an orbit with the principal quantum number n, n < m. Then

`"E"_"m" = - "me"^4/(8ε_0^2"h"^2"m"^2) and "E"_"n" = - "me"^4/(8ε_0^2"h"^2"n"^2)`

As a result, the energy radiated when an electron moves from a higher to a lower energy state is

`"E"_"m" - "E"_"n" = (-"me"^4)/(8ε_0^2"h"^2"m"^2) - ((- "me"^4)/(8ε_0^2"h"^2"n"^2))`

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

This energy is emitted as a quantum of radiation (photon) with energy hv, where v denotes the frequency of the radiation.

∴ `"E"_"m" - "E"_"n"` = hv

∴ v = `("E"_"m" - "E"_"n")/"h" = "me"^4/(8ε_0^2"h"^3)(1/"n"^2 - 1/"m"^2)`

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

The wave number, `bar"v" = 1/lambda = "v"/"c"`

∴ `bar"v" 1/lambda = "me"^4/(8ε_0^2"h"^3"c")(1/"n"^2 - 1/"m"^2) = "R"(1/"n"^2 - 1/"m"^2)` where R`(("me"^4)/(8ε_0^2"h"^3"c"))` is a constant called the Rydberg constant.

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

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

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

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

∴ `1/lambda_"B" = "R"(1/4 - 1/"m"^2)` and for the shortest wavelength line in this series,

`1/lambda_"Bs" = "R"(1/4)` as m = ∞.