An excited $He^+$ ion emits two photons in succession, with wavelengths $108.5 \, nm$ and $30.4 \, nm$, in making a transition to the ground state. The quantum number $n$, corresponding to its initial excited state is $n = ........$ (for a photon of wavelength $\lambda$, energy $E = \frac{1240 \, eV}{\lambda \, (in \, nm)}$).

If the de-Broglie wavelength of an electron in the first Bohr orbit is $\lambda$,then the minimum radial distance between the electrons in the first and second Bohr orbits is:

The possible quantum numbers for a $3d$ electron are:

If the electron in a hydrogen atom jumps from an orbit with level $n_1=2$ to an orbit with level $n_2=1$, the emitted radiation has a wavelength given by

The ratio of the areas of the electron orbits for the second excited state to the first excited state for the hydrogen atom is

Given below are two statements:
Statement $I$: In a hydrogen atom,the frequency of radiation emitted when an electron jumps from a lower energy orbit $(E_1)$ to a higher energy orbit $(E_2)$ is given as $hf = E_1 - E_2$.
Statement $II$: The jumping of an electron from a higher energy orbit $(E_2)$ to a lower energy orbit $(E_1)$ is associated with the frequency of radiation given as $f = (E_2 - E_1) / h$.
This condition is Bohr's frequency condition. In the light of the above statements,choose the correct answer from the options given below.