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:

Consider a hydrogen atom with its electron in the $n^{\text{th}}$ orbital. An electromagnetic radiation of wavelength $90 \ nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 \ eV$,then the value of $n$ is $(hc = 1242 \ eV \ nm)$.

In a hydrogen atom,the binding energy of the electron in the $n^{th}$ state is $E_n$. Then,the frequency of revolution of the electron in the $n^{th}$ orbit is:

The ratio of the ionization energy of a Bohr's hydrogen atom to that of a Bohr's hydrogen-like lithium atom is:

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:

In the first excited state of a hydrogen atom, the energy of its electron is $-3.4 \text{ eV}$. The radial distance of the electron from the hydrogen nucleus in this case is approximately: (Take $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$, $e = 1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)