Which of the following spectral series in a hydrogen atom gives a spectral line of $4860 \mathring A$?

Let the series limit for the Balmer series be $\lambda_{1}$ and the longest wavelength for the Brackett series be $\lambda_{2}$. Then $\lambda_{1}$ and $\lambda_{2}$ are related as:

Assertion : Balmer series lies in the visible region of the electromagnetic spectrum.
Reason : $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $n = 3, 4, 5, \dots$

The spectral series of the hydrogen spectrum that lies in the ultraviolet region is the

$A$ $12.5\,eV$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

The transition from the state $n = 4$ to $n = 3$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from: