If the wavelength of the first line of the Balmer series of hydrogen is 6561 Ao, the wavelength of the second line of the series should be

The wavelength of spectral line in Balmer series is given by 1λ=R 122-1n2For first line of Balmer series, n = 3⇒1λ1=R 122-132=5R36; For second line n = 4. ⇒1λ2=R 122-142=3R16∴λ2λ1=2027⇒λ1=2027×6561=4860 Ao