The product of linear momentum and angular momentum of an electron of the hydrogen atom is proportional to $n^{x}$,where $x$ is

$A$ sample of hydrogen-like atoms produces an emission spectrum consisting of $10$ wavelengths arising from all possible transitions. During this process,the maximum angular momentum change for an electron transitioning from a higher energy level to a lower energy level is:

In a Bohr atom,the total energy of an electron in the $n$-th allowed orbit is $E_n$ and its angular momentum is $J_n$. Then:

According to de-Broglie,the de-Broglie wavelength for an electron in an orbit of radius $5.3 \times 10^{-11} \ m$ of a hydrogen atom is $10^{-10} \ m$. The principal quantum number for this electron is:

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)

$A$ hydrogen atom in the ground state absorbs $ 10.2 \text{ eV} $ of energy. The orbital angular momentum of the electron is increased by: