In a hydrogen-like atom,the velocity of an electron in the second orbit is $v$. What will be the velocity of the electron in its fifth orbit?

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{2\pi}$ ($h$ is the Planck's constant). The kinetic energy $(KE)$ of the electron is (in $\text{ eV}$)

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 the electron jumps from a level $n=4$ to $n=1$,the momentum of the recoiled hydrogen atom will be

The ionisation potential of $H$-atom is $13.6 \, eV$. When it is excited from the ground state by monochromatic radiation of $970.6 \, \mathring{A}$,the number of emission lines will be (according to Bohr's theory):

$A$ 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)