$A$ hydrogen atom falls from $n^{\text{th}}$ higher energy orbit to the first energy orbit $(n=1)$. The energy released is equal to $12.75 \text{ eV}$. The $n^{\text{th}}$ orbit is:

When an electron in a hydrogen atom jumps from the third orbit to the second orbit,it emits a photon of wavelength $\lambda$. When it jumps from the fourth orbit to the third orbit,the wavelength emitted by the photon will be

Find the ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(i)$ second permitted energy level to the first level,and $(ii)$ the highest permitted energy level to the first permitted level.

$A$ free electron of $2.6 \, eV$ energy collides with a $H^+$ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. $(h = 6.6 \times 10^{-34} \, J \cdot s)$

For a certain hypothetical one-electron atom,the wavelength (in $\mathring{A}$) for the spectral lines for transition from $n = p$ to $n = 1$ is given by $\lambda = \frac{1500p^2}{p^2 - 1}$ (where $p > 1$). The ionization potential of this element must be ......$V$ (Take $hc = 12420 \text{ eV-} \mathring{A}$).

If the potential energy of a hydrogen atom in the first excited state is assumed to be zero,then the total energy of the $n = \infty$ state is,