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:

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)

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

The gyromagnetic ratio of an electron in a hydrogen atom,according to the Bohr model,is

When a hydrogen atom,initially at rest,emits a photon resulting in a transition $n = 5 \rightarrow n = 1$,its recoil speed is about

The period of revolution of an electron revolving in $n^{th}$ orbit of $H$-atom is proportional to