$A$ $12.5 \text{ eV}$ electron beam is used to bombard gaseous hydrogen at ground state. The energy level up to which the hydrogen atoms would be excited is:

What is the ratio of the total energy of an electron in a hydrogen atom in the first excited state to that in the third excited state?

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:

$A$ 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 = $ . . . . . . .

$A$ 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. $(h = 6.6 \times 10^{-34} \, J \cdot s)$

What is the total energy of an electron at an infinite distance from the nucleus in an atom?