The ratio of the velocity of an electron in the ground state of a hydrogen atom to the velocity of light (in $CGS$ units) is:

$A$ particle of mass $m$ moves in circular orbits with potential energy $V(r) = Fr$,where $F$ is a positive constant and $r$ is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle's orbit is denoted by $R$ and its speed and energy are denoted by $v$ and $E$,respectively,then for the $n^{\text{th}}$ orbit (here $h$ is the Planck's constant)-
$(A)$ $R \propto n^{2/3}$ and $v \propto n^{1/3}$
$(B)$ $R \propto n^{2/3}$ and $v \propto n^{1/3}$
$(C)$ $E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$
$(D)$ $E = 2 \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$

At which excited state of $Be^{3+}$ will the radius of the $e^{-}$ be the same as that of an $H$ atom in the ground state?

The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is

The following statements are given about the hydrogen atom:
$A$. The wavelengths of the spectral lines of the Lyman series are greater than the wavelength of the second spectral line of the Balmer series.
$B$. The orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths.

Consider an electron in the $n=3$ orbit of a hydrogen-like atom with atomic number $Z$. At absolute temperature $T$,a neutron having thermal energy $k_B T$ has the same de Broglie wavelength as that of this electron. If this temperature is given by $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$,(where $h$ is the Planck's constant,$k_B$ is the Boltzmann constant,$m_N$ is the mass of the neutron and $a_0$ is the first Bohr radius of hydrogen atom),then the value of $\alpha$ is $....$