The shortest wavelength of the H atom in the Lyman series is the longest wavelength in the Balmer series of He+ is
Explanation for the correct answer:-
Option (A)
Step 1: Calculating for
We know that,
When to be minimum that is shortest, then
For Lyman series, n=1 and for
The transition should be from
Step 2: Calculating relation between and
For the longest wavelength for the Balmer series will have minimum
Z=2 for
Hence it is the correct option.
Therefore, option (A) is the correct answer.