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:

The ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(i)$ $3^{\text{rd}}$ to $2^{\text{nd}}$ energy level and (ii) highest energy level to $3^{\text{rd}}$ level is:

Three energy levels of a hydrogen atom and the corresponding wavelengths of the emitted radiation due to different electron transitions are as shown. Then,

An electron in a hydrogen atom makes a transition from the $(n+1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to:

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......$eV$.

Assertion : Between any two given energy levels,the number of absorption transitions is always less than the number of emission transitions.
Reason : Absorption transitions start from the lowest energy level only and may end at any higher energy level. But emission transitions may start from any higher energy level and end at any energy level below it.