Question

An electron in a hydrogen atom makes transitions from orbits of higher energies to orbits of lower energies.

1. When will such transitions result in (a) Lyman (b) Balmer series?
2. Find the ratio of the longest wavelength in the Lyman series to the shortest wavelength in the Balmer series.

Answer 1

(i) Emission spectrum of Hydrogen atom: Lyman and Balmer series

• Lyman Series: When electrons will jump from a higher energy orbit to n = 1 orbit.
• Balmer Series: When electrons will jump from a higher energy orbit yo n = 2 orbits.

(ii) Longest wavelength of Lyman series:

`1/λ_"L" = "R"_"H"[1/1^2 - 1/2^2] = "R"_"H" [3/4]`

Shortest wavelength of Balmer series:

`1/λ_"B" = "R"_"H"[1/2^2 - 1/(∞^2)] = "R"_"H" [1/4]`

`1/λ_"L" = 3/4`, so `λ_"L" = 4/3` and

`1/λ_"B" = 1/4`, so `λ_"B" = 4/1`

Now,

`λ_"L"/λ_"B" = [4/3]/[4/1] = 1/3`