List of top Chemistry Questions on atomic spectra asked in BITSAT

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at:
If the series limit wavelength of Lyman series for the hydrogen atom is 912AA, then the series limit wavelength for Balmer series of hydrogen atoms is:
One of the lines in the emission spectrum of Li²+ has the same wavelength as that of the 2nd line of the Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.
The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at:
If the series limit wavelength of the Lyman series for hydrogen atom is 912AA, then the series limit wavelength for the Balmer series of hydrogen atom is:
One of the lines in the emission spectrum of \(\text{Li}^{2+}\) has the same wavelength as that of the second line of the Balmer series in hydrogen spectrum. The Balmer transition corresponds to \(n=4\rightarrow2\). If the corresponding transition in \(\text{Li}^{2+}\) is \(n=12\rightarrow x\), find the value of \(x\).
The wave number of the limiting line in Lyman series of hydrogen is \(109678\,\text{cm}^{-1}\). The wave number of the limiting line in Balmer series of He\(^+\) would be:
Hydrogen atom in ground state is excited by a monochromatic radiation of λ=975. Number of spectral lines in the resulting spectrum emitted will be