If ${\lambda _{{\text{max}}}}$ is $6563 \text{ Å}$, then the wavelength of the second line of the Balmer series will be:

The Lyman series of the hydrogen spectrum lies in which region?

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from $n = 2$ to $n = 1$ state is ...... $nm.$

If an electron in a hydrogen atom jumps from an orbit of level $n=3$ to an orbit of level $n=2$,the emitted radiation has a frequency ($R=$ Rydberg constant,$C=$ velocity of light).

If $\lambda_1$ is the wavelength of the series limit of the Lyman series,$\lambda_2$ is the wavelength of the first line of the Lyman series,and $\lambda_3$ is the series limit of the Balmer series,then the relation between $\lambda_1, \lambda_2,$ and $\lambda_3$ is:

The shortest wavelength in the Paschen series of the Hydrogen spectrum is (Rydberg constant of hydrogen $R_H = 1.097 \times 10^7 \ m^{-1}$). (in $nm$)