In the Bohr model of the hydrogen atom,the angular velocity $\omega_n$ of the electron depends on the principal quantum number $n$ as:

Potential energy between a proton and an electron is given by $U = \frac{K e^2}{3 R^3}$. Then,the radius of the Bohr orbit can be given by:

The triply ionized beryllium $(Be^{3+})$ has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given $Z = 4$ for beryllium)

When an electron jumps from the $n = 4$ level to the $n = 1$ level,the angular momentum of the electron changes by:

What is the change in angular momentum of an electron in a hydrogen atom when it transitions from the $4^{th}$ orbit to the $5^{th}$ orbit? $(h = 6.6 \times 10^{-34} \, J \cdot s)$

An electron in the hydrogen atom jumps from $n^{th}$ energy state to the ground state. The wavelength so emitted illuminates a photosensitive material having work function $2.65 \text{ eV}$. If the maximum kinetic energy of the emitted photoelectrons is $10.1 \text{ eV}$, then the value of '$n$' is