The energy equivalent to the Rydberg constant is ...... $eV$.

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:

Assuming the atom is in the ground state,the expression for the magnetic field at the nucleus in a hydrogen atom due to the circular motion of the electron is: $[\mu_0 \rightarrow \text{permeability of free space, } m \rightarrow \text{mass of electron, } \varepsilon_0 \rightarrow \text{permittivity of free space, } h \rightarrow \text{Planck's constant}]$

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:

According to Bohr's theory,the radius of an electron in an orbit described by principal quantum number $n$ and atomic number $Z$ is proportional to:

In a hydrogen-like ion,the energy difference between the $2^{\text{nd}}$ excitation state and the ground state is $108.8 \ eV$. The atomic number of the ion is: