In a hydrogen atom,the ratio of the shortest wavelength in the Balmer series to that in the Paschen series is:

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 shortest wavelength in the Balmer series of the hydrogen atom spectrum is approximately equal to (use $R_{H} = 1.097 \times 10^7 \ \text{m}^{-1}$) (in $\text{Å}$)

Find the ratio of the wavelength of the first line of the Lyman series for a doubly ionized lithium atom $(Li^{2+})$ to the wavelength of the first line of the Lyman series for deuterium $(D)$.

The longest wavelength associated with the Paschen series is: (Given $R_H = 1.097 \times 10^7 \ m^{-1}$)

The wavelength of the first line of the Balmer series of a hydrogen atom is $\lambda \ \mathring{A}$. The wavelength of the same line for a doubly ionized lithium atom $(Z = 3)$ is: