$A$ photon is emitted in transition from $n = 4$ to $n = 1$ level in a hydrogen atom. The corresponding wavelength for this transition is $......... \, nm$ (given,$h = 4 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$):

The following diagram indicates the energy levels of a certain atom. When the system moves from the $2E$ level to the $E$ level,a photon of wavelength $\lambda$ is emitted. The wavelength of the photon produced during its transition from the $\frac{4E}{3}$ level to the $E$ level is

The energy of an electron in an orbit of an $H$-atom is:

To excite the spectral line of wavelength $4960 \mathring A$ of an atom, an excitation energy of $7.7 \text{ eV}$ is required. The ground state energy of the atom is $10.5 \text{ eV}$. The energies of the two levels involved in the emission of the $4960 \mathring A$ line are (Assume $hc = 1240 \text{ eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light).

$\frac{x}{x+4}$ is the ratio of energies of photons produced due to the transition of an electron of a hydrogen atom from its $(i)$ third permitted energy level to the second level and $(ii)$ the highest permitted energy level to the second permitted level. The value of $x$ will be.

The ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(a)$ second to first energy level and $(b)$ highest energy level to the second level is: