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

At which excited state of $Be^{3+}$ will the radius of the $e^{-}$ be the same as that of an $H$ atom in the ground state?

In which of the following systems will the radius of the first orbit $(n = 1)$ be minimum?

When the electron orbiting in a hydrogen atom goes from one orbit to another orbit (principal quantum number $= n$),the de-Broglie wavelength $(\lambda)$ associated with it is related to $n$ as:

In the Bohr model of a hydrogen-like atom,the force between the nucleus and the electron is modified as $F = \frac{e^2}{4\pi \varepsilon_0} \left( \frac{1}{r^2} + \frac{\beta}{r^3} \right)$,where $\beta$ is a constant. For this atom,the radius of the $n^{th}$ orbit in terms of the Bohr radius $\left( a_0 = \frac{\varepsilon_0 h^2}{m \pi e^2} \right)$ is:

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: