For the complete Balmer series of hydrogen, let λₘₐₓ be the longest wavelength and λₘᵢₙ be the series-limit shortest wavelength. If λₘₐₓ ÷ λₘᵢₙ = N ÷ 5, find N.
Physics · Atomic Structure and X-rays · atomic structure and bohr atomic model
JEE Mains NEET JEE Advanced

Question

For the complete Balmer series of hydrogen, let λₘₐₓ be the longest wavelength and λₘᵢₙ be the series-limit shortest wavelength. If λₘₐₓ ÷ λₘᵢₙ = N ÷ 5, find N.

G1 Physics · Accepted Answer

For the Balmer series, all transitions end at n = 2. The longest wavelength is produced by the smallest Balmer energy difference, from n = 3 to n = 2: 1 ÷ λₘₐₓ = R[(1 ÷ 2²)−(1 ÷ 3²)] = R(1 ÷ 4−1 ÷ 9) = 5R ÷ 36. Thus, λₘₐₓ = 36 ÷ (5R). At the Balmer series limit: 1 ÷ λₘᵢₙ = R ÷ 4, so λₘᵢₙ = 4 ÷ R. Therefore, λₘₐₓ ÷ λₘᵢₙ = [36 ÷ (5R)] ÷ [4 ÷ R] = 9 ÷ 5. Hence N = 9.