The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

Solution

Physics sir · 3 steps

Shortest wavelength means the electron falls from infinity, so only the lower level decides it.

1. Shortest means largest jump. Wavelength is smallest when the energy released is largest. In any series that happens when the electron falls from n₂ = ∞ down to the series' lower level n₁.

   1/λ = R(1/n₁²-1/∞²) = R/n₁² ⇒ λₘᵢₙ = n₁²/R

2. Apply to each series. Balmer ends on n₁ = 2, so λ_B = 4/R. Lyman ends on n₁ = 1, so λ_L = 1/R.
3. So. λ_B : λ_L = 4/R : 1/R = 4 : 1, which is option (d).

The idea to keep. The shortest wavelength of any series is its series limit, λₘᵢₙ = n₁²/R. Compare series limits by comparing n₁² directly.

Answer from the NTA final answer key. Solution by PrepWiser.