Question

Calculate the longest wavelength of the Balmer series for the H atom.

Answer 1

Formula: `1/λ =  R_H(1/n^2 - 1/m^2)` Here n < m

where, λ is the wavelength,

R → Rydberg’s constant = 1.097 × 10⁷ m⁻¹

For the longest wavelength of the Balmer series,

n = 2, m = 3

∴ `1/λ_L = 1.097 xx 10^7(1/2^2 - 1/3^2)`

∴ `1/λ_L = 1.097 xx 10^7(1/4 - 1/9)`

∴ `1/λ_L = 1.097 xx 10^7(5/36)`

∴ `λ_L = (36 xx 10^-7)/(5 xx 1.097)`

∴ λ_L = 6563 × 10⁻¹⁰

∴ λ_L = 6563 Å