$A$ hydrogen sample is prepared in a particular excited state $A$ of quantum number $n_A=3$. The ground state energy of the hydrogen atom is $-|E|$. Photons of energy $\frac{|E|}{12}$ are absorbed by the sample,which results in the excitation of some electrons to an excited state $B$ of quantum number $n_B$. The value of $n_B$ is:

The energy of a hydrogen atom in its ground state is $-13.6 \, eV$. The energy of the level corresponding to the quantum number $n = 5$ is ...... $eV$.

Out of the following transitions in a hydrogen atom,identify the transition which emits photons of highest frequency.

The figure shows the energy levels of a certain atom. When the electron de-excites from $3E$ to $E$,an electromagnetic wave of wavelength $\lambda$ is emitted. What is the wavelength of the electromagnetic wave emitted when the electron de-excites from $\frac{5E}{3}$ to $E$?

The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. When its electron is in the first excited state,its excitation energy is:

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.