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:

The ground state energy of a hydrogen atom is $-13.6 \; eV$. What are the kinetic and potential energies of the electron in this state?

Consider a hydrogen-like atom whose energy in the $n^{th}$ excited state is given by $E_n = - \frac{13.6 Z^2}{n^2}$. When this excited atom makes a transition from an excited state to the ground state, the most energetic photons have energy $E_{max} = 52.224 \ eV$ and the least energetic photons have energy $E_{min} = 1.224 \ eV$. The atomic number of the atom is:

$A$ hydrogen atom in its ground state absorbs $10.2 \ eV$ of energy. The orbital angular momentum is increased by (Given Planck constant $h = 6.6 \times 10^{-34} \ J \cdot s$)

An electron in an excited state of $Li^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de-Broglie wavelength of the electron in this state is $p \pi a_{0}$ (where, $a_{0} = \text{Bohr radius}$). The value of $p$ is

In a hydrogen atom,if $V_n$ and $V_p$ are the orbital velocities in the $n^{\text{th}}$ and $p^{\text{th}}$ orbits respectively,then the ratio $V_p : V_n$ is: