The ratio of acceleration of the electron in a singly ionized Helium atom $(He^+)$ to that of a Hydrogen atom $(H)$ (both in the ground state) is:

In a hydrogen atom,the electron and proton are bound at a distance of about $0.53 \; \mathring{A}$.
$(a)$ Estimate the potential energy of the system in $eV$,taking the zero of the potential energy at infinite separation of the electron from the proton.
$(b)$ What is the minimum work required to free the electron,given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in $(a)$?
$(c)$ What are the answers to $(a)$ and $(b)$ above if the zero of potential energy is taken at $1.06 \; \mathring{A}$ separation?

An electron in the hydrogen atom initially in the fourth excited state makes a transition to $n^{\text{th}}$ energy state by emitting a photon of energy $2.86 \ eV$. The integer value of $n$ will be . . . . . . .

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

If the first excitation potential of a hydrogen-like atom is $V \, \text{volt}$, then the ionization energy of this atom will be

The Bohr model of atoms: