According to Bohr's theory,the moment of momentum of an electron revolving in the $4^{\text{th}}$ orbit of a hydrogen atom is:

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

Hydrogen $(_1H^1)$,Deuterium $(_1H^2)$,singly ionized Helium $(_2He^4)^+$,and doubly ionized Lithium $(_3^6Li)^{++}$ all have one electron around the nucleus. Consider an electron transition from $n = 2$ to $n = 1$. If the wavelengths of emitted radiation are $\lambda_1, \lambda_2, \lambda_3$,and $\lambda_4$ respectively,then approximately which one of the following is correct?

Using the formula for the radius of the $n^{th}$ orbit $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,derive an expression for the total energy of an electron in the $n^{th}$ Bohr orbit.

An electron is revolving in a circular orbit of radius '$r$' in a hydrogen atom. Using Bohr's theory, the angular momentum of the electron is (where '$M$' = magnetic dipole moment,'$m$' = mass of electron,'$e$' = charge of electron).

The ionization energy of hydrogen is $13.6 \, eV$. If $h = 6.6 \times 10^{-34} \, J \cdot s$,the order of magnitude of the Rydberg constant $R$ will be: