The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
(2) 4
The last line (series limit) of a series corresponds to the transition from ni = .
For the Balmer series (nf = 2), 1 B = R ( 122 - 1 2 ) = R4, so
B = 4R
For the Lyman series (nf = 1), 1 L = R ( 112 - 1 2 ) = R, so
L = 1R
∴ B L = 4/R1/R = 4
Answer: (2)
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