Which of the following spectral series of the hydrogen atom lies in the visible range of the electromagnetic spectrum?

For a hydrogen atom,$\lambda_1$ and $\lambda_2$ are the wavelengths corresponding to the transitions $1$ and $2$ respectively,as shown in the figure. The ratio of $\lambda_1$ and $\lambda_2$ is $\frac{x}{32}$. The value of $x$ 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.

If the electron in a hydrogen atom transits from the second orbit to the first orbit,in what region would the emitted radiation be?

In a hydrogen spectrum,let $\lambda$ be the wavelength of the first transition line of the Lyman series. The wavelength difference between the $3^{\text{rd}}$ transition line of the Paschen series and the $2^{\text{nd}}$ transition line of the Balmer series is $a\lambda$,where $a = ........$

The second line of the Balmer series has a wavelength of $4861 Å$. The wavelength of the first line of the Balmer series is: (in $Å$)