Questions
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=527Similar Questions
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