The wavelength of the first spectral line of the Lyman series of hydrogen spectrum is

Step-by-step solution

1 λ = R 1 n 1 2 − 1 n 2 2 ⋅ Z 2 where, R = Rydbergs constant (Also written is R H ) Z = atomic number Since the question is asking for 1 st line of Lyman series therefore n 1 = 1 n 2 = 2 Since the electron is de-exited from 1 st exited sate i . e . n = 2 to ground sate i . e . n = 1 for first line of Lyman series. 1 λ = R 1 1 2 − 1 2 2 ⋅ 1 2 Since the atomic number of Hydrogen is 1 By doing the math, we get the wavelength as λ = 4 3 ⋅ 912   Å Since 1 R = 912 &