The radius of the first (lowest) orbit of the hydrogen atom is $a_0$. The radius of the second (next higher) orbit will be (in $a_0$)

If $m$ is the mass of an electron,$v$ is its velocity,and $r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's postulate,the kinetic energy $K = \frac{1}{2}mv^2$ of the electron in the $C.G.S.$ system is equal to:

The magnetic moment $(m_{orb})$ of a revolving electron around the nucleus varies with the principal quantum number $(n)$ as

$13.6 \text{ eV}$ energy is required to separate a hydrogen atom into a proton and an electron. If the orbital radius of an electron in a hydrogen atom is $5.3 \times 10^{-11} \text{ m}$, then the velocity of the electron is . . . . . . .

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

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