Question

The wavenumber of the shortest wavelength transition in the balmer series of atomic hydrogen is

Answer 1

The shortest wavelength transition in the Balmer series of atomic hydrogen corresponds to the transition of n = 2 to n = ∞

So, n = 2 → n = ∞.

Therefore, n₁ = 2, n₂ = ∞

v = RH (1/n²₁ - 1/n²₂) = (109677 cm⁻¹) (1/2² - 1/∞²) = 27419.25 cm⁻¹