If $13.6 \ eV$ energy is required to ionize the hydrogen atom,then the energy required to remove an electron from $n=2$ is (in $eV$)

An electron in a hydrogen atom makes a transition from the $(n+1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to:

The ionisation potential of a hydrogen atom is $13.6 \, V$. The energy required to remove an electron in the $n = 2$ state of the hydrogen atom is.....$eV$.

$A$ hydrogen atom is bombarded with electrons accelerated through a potential difference of $V$, which causes excitation of hydrogen atoms. If the experiment is performed at $T = 0 \,K$, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{\alpha}{10} \,V$, where $\alpha = $ . . . . . . .

$A$ monochromatic light is incident on a hydrogen sample in the ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is $x \times 10^{15} \ Hz$. The value of $x$ is (Given $h = 4.25 \times 10^{-15} \ eVs$)

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).