Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

detailed solution

Correct option is C

The wavelength of different spectral lines of Lyman series is given by1λL=R1l2−1n2   where  n=2,3,4,....where subscript L refers to Lyman. For longest wavelength, n = 2∴   1λLlongest=R112−122=34R           −−−(i)The wavelength of different spectral series of Balmer series is given by1λB=R122−1n2  where  n=3,4,5,....where subscript B refers to Balmer. For longest wavelength, n= 3∴  1λBlongest=R122−132=R14−19=5R36   −−−(ii)Divide (ii) by (i), we getλLlongestλBlongest=5R36×43R=527