The four lowest energy levels of an $H$-atom are shown in the figure. The number of possible emission lines would 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:

$A$ photon of wavelength $\lambda$ is absorbed by an electron confined to a box of length $L = \sqrt{(35 h \lambda / 8 m c)}$. As a result,the electron makes a transition from state $k=1$ to the state $n$. Subsequently,the electron transitions from the state $n$ to the state $m$ by emitting a photon of wavelength $\lambda^{\prime} = 1.75 \lambda$. Then,

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

The energy of a hydrogen atom in its ground state is $-13.6 \,eV$. The energy of the level corresponding to $n=7$ is ...... $eV$.

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