$A$ hydrogen atom emits a photon corresponding to an electron transition from $n=5$ to $n=1$. If $R$ is the Rydberg constant, then the wavelength of the emitted photon is

$A$ doubly ionised $Li$ atom is excited from its ground state $(n = 1)$ to $n = 3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratios $\lambda_{32}/\lambda_{31}$ and $\lambda_{21}/\lambda_{31}$ are,respectively:

The frequencies for the series limit of the Balmer and Paschen series are $v_1$ and $v_3$ respectively. If the frequency of the first line of the Balmer series is $v_2$,then the relation between $v_1, v_2,$ and $v_3$ is:

The shortest wavelength in the Lyman series is $912 \ \text{Å}$. Then, the longest wavelength in the series must be (in $\text{Å}$)

The wavelengths corresponding to the first four spectral lines of the Lyman series of the $H$-atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}$,and $951.4 \, \mathring{A}$. Now,consider a deuterium atom instead of a hydrogen atom. Given the mass of the hydrogen atom is $1.6725 \times 10^{-27} \, kg$,the mass of the deuterium atom is $3.3374 \times 10^{-27} \, kg$,and the mass of the electron is $9.109 \times 10^{-31} \, kg$,calculate the percentage change in the wavelength of the first spectral line of the Lyman series for the deuterium atom relative to the hydrogen atom.

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