The radius of a certain orbit of a hydrogen atom is $8.48 \mathring{A}$. If the energy of the electron in this orbit is $E/x$,then $x = . . . .$
(Given $a_0 = 0.529 \mathring{A}$,$E =$ energy of the electron in the ground state)

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the $2^{nd}$ and $4^{th}$ orbits of hydrogen atom. (in $32$ : $1$)

The electron in a hydrogen atom makes a transition from the $M$ shell to the $L$ shell. The ratio of the magnitudes of the initial to the final centripetal acceleration of the electron is:

If $m$ is the mass of an electron,$v$ is its velocity,$r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's first postulate,the kinetic energy of the electron is (where $K = 1 / 4 \pi \epsilon_0$):

Hydrogen $(H)$,deuterium $(D)$,singly ionized helium $(He^+)$ and doubly ionized lithium $(Li^{2+})$ all have one electron around the nucleus. Consider the $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ respectively. Then approximately:

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: