The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen-like ion. The atomic number Z of hydrogen-like ion is
- 3
- 4
- 1
- 2
Answer
(4) 2
Solution
For a hydrogen-like ion: 1 = RZ2 ( 1n12 - 1n22 )
First Lyman line of H (n = 2 1, Z = 1):
1 = R (1 - 14 ) = 3R4
Second Balmer line of the ion (n = 4 2):
1 = RZ2 ( 14 - 116 ) = 3RZ216
Equating:
3RZ216 = 3R4
Z2 = 4
Z = 2
Answer: (4)