When an electron is excited from its $4^{\text{th}}$ orbit to its $5^{\text{th}}$ stationary orbit,the change in the angular momentum of the electron is approximately (Planck's constant $= h = 6.63 \times 10^{-34} \text{ J s}$)

Assertion : Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate.
Reason : According to classical physics all moving electrons radiate.

Speed of electron in its $1^{st}$ Bohr's orbit is given by $2.18 \times 10^6 \ m/s$. If the time period of electron in $n^{th}$ orbit is measured as $4.10 \ fs$,the value of $n$ is

The triply ionized beryllium $(Be^{3+})$ has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given $Z = 4$ for beryllium)

The radius of the first permitted Bohr orbit for the electron in a hydrogen atom equals $0.53 \mathring{A}$ and its ground state energy equals $-13.6 \; eV$. If the electron in the hydrogen atom is replaced by a muon $(\mu^{-})$ [charge same as electron and mass $207 m_{e}$], the first Bohr radius and ground state energy will be:

$A$ 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?
$(A)$ $r^2 = n\hbar \sqrt{\frac{1}{mk}}$
$(B)$ $v^2 = n\hbar \sqrt{\frac{k}{m^3}}$
$(C)$ $\frac{L}{mr^2} = \sqrt{\frac{k}{m}}$
$(D)$ $E = \frac{n\hbar}{2} \sqrt{\frac{k}{m}}$