If the series limit of the Lyman series for a Hydrogen atom is equal to the series limit of the Balmer series for a hydrogen-like atom,then the atomic number of this hydrogen-like atom will be:

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 $R = 1.097 \times 10^7 \ m^{-1}$)

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

The shortest wavelength in the Lyman series is $91.2 \, nm$. The largest wavelength of the series is.....$nm$

In hydrogen atom spectra,if the ratio of wavelengths corresponding to the first line of the Lyman series and the first line of the Balmer series is $9 \alpha$,the value of $\alpha$ is

If an electron in a hydrogen atom jumps from an orbit of level $n=3$ to an orbit of level $n=2$,the emitted radiation has a frequency ($R=$ Rydberg constant,$C=$ velocity of light).