$A$ hydrogen atom in its ground state absorbs $10.2 \; eV$ of energy. The angular momentum of the electron of the hydrogen atom will increase by the value of ............... $\times 10^{-34} \; J \cdot s$ (Given,Planck's constant $h = 6.6 \times 10^{-34} \; J \cdot s$)

$A$ particle of mass $m$ is moving in a circular orbit under the influence of the central force $F(r) = -kr$,corresponding to the potential energy $V(r) = \frac{1}{2}kr^2$,where $k$ is a positive force constant and $r$ is the radial distance from the origin. According to Bohr's quantization rule,the angular momentum of the particle is given by $L = n\hbar$,where $\hbar = \frac{h}{2\pi}$,$h$ is Planck's constant,and $n$ is a positive integer. If $v$ and $E$ are the speed and total energy of the particle,respectively,then which of the following expression$(s)$ is(are) correct?
$(A)$ $r^2 = n\hbar \sqrt{\frac{1}{mk}}$
$(B)$ $v^2 = n\hbar \sqrt{\frac{k}{m^3}}$
$(C)$ $\frac{L}{mr^2} = \sqrt{\frac{k}{m}}$
$(D)$ $E = \frac{n\hbar}{2} \sqrt{\frac{k}{m}}$

An electron in an excited state of $Li^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de-Broglie wavelength of the electron in this state is $p \pi a_{0}$ (where, $a_{0} = \text{Bohr radius}$). The value of $p$ is

An orbital electron in the ground state of hydrogen has a magnetic moment $\mu_1$. This orbital electron is excited to the $3^{rd}$ excited state by some energy transfer to the hydrogen atom. If the new magnetic moment of the electron is $\mu_2$,then:

The number of revolutions per second made by an electron in the first Bohr orbit of a hydrogen atom is ($h$ = Planck's constant, $m$ = mass of electron, $r$ = radius of the orbit).

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: