Every series of the hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to $18752 \mathring{A}$ is:
(Rydberg constant $R = 1.097 \times 10^7 \ m^{-1}$)

The wavelength of radiation emitted is $\lambda_0$ when an electron jumps from the second excited state to the first excited state of a hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom,the wavelength of the radiation emitted will be $\frac{20}{x} \lambda_0$. The value of $x$ is

In a hydrogen-like atom,when an electron transits from energy state $n=5$ to $n=2$,a photon of wavelength $434 \ nm$ is emitted. What will be the wavelength (in $nm$) of the photon emitted when the transition occurs from energy state $n=4$ to $n=2$?

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

In the hydrogen spectrum,the shortest and longest wavelengths of the Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant $R$ of hydrogen is:

Explain emission line spectra and absorption spectra.