As per the Bohr model,the minimum energy (in $eV$) required to remove an electron from the ground state of a doubly ionized $Li$ atom $(Z = 3)$ is:

An electron has a mass of $9.1 \times 10^{-31} \ kg$. It revolves around the nucleus in a circular orbit of radius $0.529 \times 10^{-10} \ m$ at a speed of $2.2 \times 10^6 \ m/s$. The magnitude of its linear momentum in this motion is:

In a hydrogen atom,the binding energy of the electron in the $n^{th}$ state is $E_n$. Then,the frequency of revolution of the electron in the $n^{th}$ orbit is:

An electron in an excited state of $Li^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de-Broglie wavelength of the electron in this state is $p \pi a_{0}$ (where, $a_{0} = \text{Bohr radius}$). The value of $p$ is

The energy of a hydrogen atom in the ground state is $-13.6 \, eV$. The energy of the $He^+$ ion in the first excited state will be .... $eV$.

Magnetic moment due to the motion of the electron in the $n$th energy state of a hydrogen atom is proportional to ..........