The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :
- 1 : 2
- 1 : 4
- 2 : 1
- 4 : 1
Answer
(4) 4 : 1
Solution
1 = R ( 1n12 - 1n22 ); shortest n2 =
Balmer (n1 = 2): B = 4R
Lyman (n1 = 1): L = 1R
B : L = 4 : 1
Answer: (4)