In the Bohr model of the $H$-atom,the kinetic energy of an electron in any orbit depends on the principal quantum number $n$ as:

Excitation energy of a hydrogen-like ion in its first excitation state is $40.8 \, eV$. Energy needed to remove the electron from the ion in the ground state is ........ $eV$.

In the $n^{\text{th}}$ Bohr orbit,the ratio of the kinetic energy of an electron to the total energy of it is:

The magnitude of total energy and angular momentum of an electron in the $n^{th}$ orbit of a Bohr atom is denoted by $E_{n}$ and $L_{n}$ respectively. Then

In a hypothetical Bohr hydrogen atom, if the mass of the electron is doubled, then the energy of the electron in the first orbit is: (in $\text{ eV}$)

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 $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)