Question

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer 1

Wave number \(\bar{V}\) = 109677 \(\big(\frac{1}{n_1^2}-\frac{1}{n_2^2}\big)\)

n₁ = 2

n₂ = 3 For Balmer series

\(\therefore\) \(\bar{V}\) = 109677 \(\Big(\frac{1}{(2)^2}-\frac{1}{(3)^2}\Big)\)

= 109677 \(\Big(\frac{1}{4}-\frac{1}{9}\Big)\)

= 109677 \(\Big(\frac{9-4}{36}\Big)\)

= 109677 x \(\frac{5}{36}\)

= 1.523 × 10⁴ cm⁻¹