The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. The potential energy of the electron in this state is: (in $\text{ eV}$)

Let $v_{n}$ and $E_{n}$ be the respective speed and energy of an electron in the $n$th orbit of radius $r_{n}$, in a hydrogen atom, as predicted by Bohr's model. Then:

Electrons in a certain energy level $n=n_{1}$ can emit $3$ spectral lines. When they are in another energy level $n=n_{2}$,they can emit $6$ spectral lines. The orbital speed of the electrons in these orbits are in the ratio:

The angular momentum of an electron in the hydrogen atom is $\frac{3h}{2\pi}$. Here $h$ is Planck's constant. The kinetic energy of this electron is.....$eV$

Hydrogen $(_1H^1)$,Deuterium $(_1H^2)$,singly ionized Helium $(_2He^4)^+$,and doubly ionized Lithium $(_3^6Li)^{++}$ all have one electron around the nucleus. Consider an electron transition from $n = 2$ to $n = 1$. If the wavelengths of emitted radiation are $\lambda_1, \lambda_2, \lambda_3$,and $\lambda_4$ respectively,then approximately which one of the following is correct?

The time of revolution of an electron around a nucleus of charge $Ze$ in the $n^{th}$ Bohr orbit is directly proportional to