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 $\text{ nm}$)

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} = $

$A$ hydrogen atom absorbs radiation of wavelength $975 \, \mathring{A}$ and transitions from the ground state to an excited state. How many spectral lines are possible in the emission spectrum?

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

$A$ hydrogen atom is excited from the ground state to the energy level $n = 3$. According to Bohr's model,the number of spectral lines emitted is:

Using the Rydberg formula,calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.