In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is

Step-by-step solution

When an atom comes down from some higher energy level to the first energy level then emitted lines form of Lyman series. 1 λ L = R 1 1 2 - 1 n 2 where R is Rydberg's constant. When an atom comes from higher energy level to the second level, then Balmer series are obtained. 1 λ B = R 1 2 2 - 1 n 2 For maximum wavelength n ⁡ = 2 , 1 λ L = R 1 - 1 ( 2 ) 2 = R 1 - 1 4 = 3 R 4 … . ( i ) n = 3 , 1 λ B = R 1 ( 2 ) 2 - 1 ( 3 ) 2 = R 5 36 … ( i i ) Dividing Eq. (ii) by Eq. (i), we get λ L λ B = 5 27