The ratio of energies of photons produced due to the transition of an electron of a hydrogen atom from its $(i)$ second to first energy level and $(ii)$ highest energy level to $2^{nd}$ level is respectively: (in $: 1$)

The energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{-13.6}{n^2} \; eV$. The energy of a photon emitted when the electron jumps from the $n = 3$ state to the $n = 2$ state of hydrogen is approximately......$eV$.

In a hydrogen atom, the energy of an electron in the first and third orbits is $E_1$ and $E_3$ respectively. If $E_3 = x E_1$, then the value of $x$ will be:

An electron jumps from the $4th$ orbit to the $2nd$ orbit of a hydrogen atom. Given the Rydberg constant $R = 10^5 \text{ cm}^{-1}$. The frequency in $\text{Hz}$ of the emitted radiation will be:

What is an emission line?

The energy (in $eV$) required to excite an electron from $n=2$ to $n=4$ state in a hydrogen atom is: