The wavelength of light for the least energetic photons emitted in the Lyman series of the hydrogen spectrum is nearly. [Take $hc = 1240 \text{ eV-nm}$, change in energy of the levels $= 10.2 \text{ eV}$] (in $\text{ nm}$)

$A$ hydrogen atom is excited from the ground state to a state with principal quantum number $n = 4$. The number of spectral lines emitted in the emission spectrum is:

Which of the following spectral series in a hydrogen atom gives a spectral line of $4860 \mathring A$?

If $\Delta \lambda_L$ is the difference between the shortest and longest wavelengths of the Lyman series and $\Delta \lambda_B$ is the difference between the shortest and longest wavelengths of the Balmer series,then $\frac{\Delta \lambda_B}{\Delta \lambda_L} = $

In a hydrogen atom,a transition from $n = 4$ to $n = 3$ results in the emission of an infrared photon. Which of the following transitions would also result in the emission of an infrared photon?

The ratio of the maximum wavelengths of the Lyman and Balmer series in the hydrogen spectrum is ........