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

The energy of a hydrogen atom in the ground state is $-13.6 \, eV$. The energy of the $He^+$ ion in the first excited state will be .... $eV$.

In a hydrogen atom and a $Li^{2+}$ ion,the electron is in the second excited state. If $l_{H}$ and $l_{Li}$ are the angular momenta of the electrons and $E_H$ and $E_{Li}$ are their respective energies,then:

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 $R_0 = 1 \,\mathring{A}$.

In a hydrogen atom,an electron makes a transition from the $n=4$ to the $n=1$ energy level. The recoil momentum of the $H$ atom will be:

$A$ particle of mass $m$ moves around the origin in a potential $\frac{1}{2} m \omega^{2} r^{2}$,where $r$ is the distance from the origin. Applying the Bohr's model in this case,the radius of the particle in its $n$th orbit in terms of $a=\sqrt{\frac{h}{2 \pi m \omega}}$ is