Q.
Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
729
b
931
c
527
d
323
answer is C.
(Unlock A.I Detailed Solution for FREE)
Detailed Solution
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
Watch 3-min video & get full concept clarity