What is the maximum wavelength of the $H_{\alpha}$ line in the Balmer series of the hydrogen atom (in $nm$)?

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 terms of Rydberg constant $R,$ the shortest wavelength in the Balmer series of the Hydrogen atom spectrum will be:

What is the ratio of the wavelengths of radiations emitted when an electron in a hydrogen atom jumps from the fourth orbit to the second orbit and from the third orbit to the second orbit?

Let $F_1$ be the frequency of the second line of the Lyman series and $F_2$ be the frequency of the first line of the Balmer series. Then,the frequency of the first line of the Lyman series is given by:

For the wavelength of visible radiation of the Hydrogen spectrum, Balmer gave an equation as $\lambda = \frac{xm^2}{m^2 - 4}$, where $m$ is an integer value. The value of $x$ in terms of Rydberg's constant $R$ is