According to Bohr's theory,the moment of momentum of an electron revolving in the $4^{\text{th}}$ orbit of a hydrogen atom is:

In a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is $r_n$. The graph between $\log \left( \frac{r_n}{r_1} \right)$ and $\log n$ will be:

The ratio of the speed of an electron in the ground state of the Bohr's first orbit of a hydrogen atom to the velocity of light in air is:

In the Bohr model of the hydrogen atom,the centripetal force is provided by the Coulomb attraction between the proton and the electron. If $r_0$ is the radius of the ground state orbit,$m$ is the mass,$e$ is the charge on the electron,and $\varepsilon_0$ is the permittivity of vacuum,the speed of the electron is:

For $He^{+}$, a transition takes place from the orbit of radius $105.8 \ pm$ to the orbit of radius $26.45 \ pm$. The wavelength (in $nm$) of the emitted photon during the transition is. . . . .
[Use: Bohr radius, $a_0=52.9 \ pm$; Rydberg constant, $R_H=2.2 \times 10^{-18} \ J$; Planck's constant, $h=6.6 \times 10^{-34} \ J \ s$; Speed of light, $c=3 \times 10^8 \ m \ s^{-1}$]

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ separated by a distance $r$. Applying the Bohr's quantization rule for angular momentum,calculate its rotational kinetic energy. It is given by the formula: