If the wavelength of the first line of the Balmer series of hydrogen is 6561A0, find the wavelength of the second line of the series.

The wavelength of spectral line in Balmer series is given by 1λ=R[122−1n2] For first line of Balmer series, n=3⇒1λ1=R[122−132]=5R36 For second line n=4⇒1λ2=R[122−142]=3R16∴λ2λ1=2027⇒λ2=2027×6561=4860Å