What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum (in $\text{ nm}$)? (Take $hc = 1240 \text{ eV nm}$)

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the third member of the Lyman series and the first member of the Paschen series respectively,then the value of $\lambda_{1} : \lambda_{2}$ is

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

The ionization energy of the electron in the hydrogen atom in its ground state is $13.6 \text{ eV}$. The atoms are excited to higher energy levels to emit radiations of $6$ wavelengths. The maximum wavelength of the emitted radiation corresponds to the transition between:

The ratio of the wavelengths of the first Lyman line and the second Balmer line of the hydrogen atom is

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