For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is
- 5 : 27
- 27 : 5
- 3 : 4
- 5 : 36
Answer
(1) 5 : 27
Solution
1 = R ( 1n12 - 1n22 ); the largest wavelength comes from the smallest energy gap.
Lyman (2 1): 1 L = R (1 - 14 ) = 3R4
Balmer (3 2): 1 B = R ( 14 - 19 ) = 5R36
⇒ L B = 5R/363R/4 = 527
∴ L : B = 5 : 27
Answer: (1)