If the speed of an electron of a hydrogen atom in the ground state is $2.2 \times 10^6 \ m/s$,then its speed in the third excited state will be

In the Bohr model of the hydrogen atom,the electrostatic force on the electron depends on the principal quantum number $n$ as:

The ionisation energy of a hydrogen atom is $13.6 \; eV$. The ionisation energy of a singly ionised helium atom would be ....... $eV$.

If the energy of a hydrogen atom in the $n^{th}$ orbit is ${E_n}$,then the energy in the $n^{th}$ orbit of a singly ionized helium atom will be:

Assertion $(A)$: The magnetic moment $(\mu)$ of an electron revolving around the nucleus decreases with increasing principal quantum number $(n)$.
Reason $(R)$: Magnetic moment of the revolving electron,$\mu \propto n$.

What is the minimum energy that must be given to a $H$ atom in the ground state so that it can emit an $H_{\gamma}$ line in the Balmer series? If the angular momentum of the system is conserved,what would be the angular momentum of such an $H_{\gamma}$ photon?