When the wave of a hydrogen atom comes from infinity into the first orbit,then the value of the wave number is......$cm^{-1}$

The smallest wavelength of the Lyman series is $91 \ nm$. The difference between the largest wavelengths of the Paschen and Balmer series is nearly . . . . . . $nm$.

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 $n=2$?

Assertion: In the Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
Reason: The Lyman series constitutes spectral lines corresponding to transitions from higher energy levels to the ground state of the hydrogen atom.

In the spectrum of hydrogen,the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

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.