Wavelength of first line in Lyman series is λ . The wavelength of first line in Balmer series is

Step-by-step solution

According to Bohr, the wavelength emitted when an electron jumps from n 1 t h to n 2 t h orbit is E = h c λ = E 2 – E 1 1 λ = R 1 n 1 2 - 1 n 2 2 For first-line in Lyman series 1 λ L = R 1 1 2 - 1 2 2 = 3 R 4 ...(i) For first-line in Balmer series, 1 λ B = R 1 2 2 - 1 3 2 = 5 R 36 ...(ii) From Eqs. (i) and (ii) ∴ λ B λ L = 3 R 4 × 36 5 R = 27 5 ∴ λ B = 27 5 λ ( ∵ λ L = λ )