The energy of an electron in the fourth excited state of the hydrogen atom is

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.

An electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases 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 = $ . . . . . . .

An electron with kinetic energy $E$ collides with a hydrogen atom in the ground state. The collision will be elastic

The ground state energy of the hydrogen atom is $-13.6 \ eV$. The kinetic and potential energy of the electron in the second excited state are,respectively: