The electrostatic potential energy of the electron in an orbit of hydrogen is $-6.8 \ eV$. The speed of the electron in this orbit is ($C$ is the speed of light in vacuum).

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance $r$. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization,its energy will be given by: ($n$ is an integer)

The Bohr model for the $H$-atom relies on Coulomb's law of electrostatics. Coulomb's law has not been directly verified for very short distances of the order of $\mathring{A}$. Suppose Coulomb's law between two opposite charges $+q_1$ and $-q_2$ is modified to $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{r^2} \right)$ for $r \ge R_0$ and $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{R_0^{2-\epsilon} r^{\epsilon}} \right)$ for $r < R_0$. Calculate the ground state energy of an $H$-atom,given $\epsilon = 0.1$ and $R_0 = 1 \,\mathring{A}$.

When a hydrogen atom absorbs a photon of wavelength $60 \ nm$,the atom undergoes photo-ionization. What will be the maximum kinetic energy of the emitted electron in $eV$?

An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocities of the electron in these two orbits is

In a hydrogen atom, the electron makes $6.6 \times 10^{15}$ revolutions per second around the nucleus in an orbit of radius $0.5 \times 10^{-10} \, m$. This is equivalent to a current of nearly: