Total energy of an electron in an excited state of a hydrogen atom is $-3.4 \text{ eV}$. Find the kinetic and potential energy of the electron in this state.

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

The potential energy of the orbital electron in the ground state of a hydrogen atom is $-E$. What is its kinetic energy?

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:

The number of spectral lines emitted by atomic hydrogen that is in the $4^{\text{th}}$ energy level is

In a hydrogen-like atom,when an electron jumps from the $M$-shell to the $L$-shell,the wavelength of the emitted radiation is $\lambda$. If an electron jumps from the $N$-shell to the $L$-shell,the wavelength of the emitted radiation will be: