The shortest wavelength of the Lyman series of a hydrogen atom is equal to the shortest wavelength of the Balmer series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is equal to:

In an $H$-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 of the photon emitted when the transition occurs from energy state $n=4$ to $n=2$?

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 $..........$

If ${\lambda _{\max }}$ is $6563\,\mathring{A}$ for the Balmer series of a particular atom,then the wavelength of the second line for the Balmer series will be:

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

The wavelength of the first line in the Balmer series in the hydrogen spectrum is $\lambda$. What is the wavelength of the second line in the same series?