Taking into account the motion of the nucleus of a hydrogen atom, choose the correct statements. Take  $\mu =\frac{M{m}_e}{m_e+M};M$: Mass of proton,  $h=h/2\pi ;h$ : planks constant

Given that  M is the mass of the nucleus (proton) and m  the mass of electron. Proton and electron both revolve about their centre of mass (COM) with same angular velocity $\left(\omega \right)$   but different linear speeds. Let  $r_1$ and $r_2$  be the distance of COM from proton and electron.
Let r  be the distance between the proton and the electron. Then
$$\begin{gathered} M{r}_1=m{r}_2 \\ {r}_1+r_2=r \end{gathered}$$

$\therefore r_1=\frac{mr}{M+m} and r_2=\frac{Mr}{M+m}$

Centripetal force to electron is provided by the electrostatic force.
 $\mu r^3{\omega }^2=\frac{e^2}{4\pi {\epsilon }_0}\mathrm{...}\left(1\right)$
Moment of inertia of the atom about COM is :
 $I=M{r}_1^2+m{r}_2^2$
or  $I=\left(\frac{Mm}{M+m}\right)r^2=\mu r^2\mathrm{...}\left(2\right)$
According to Bohr’s theory  $I\omega =nh$ where  $(h=\frac{h}{2\pi })$
or  $\mu r^2\omega =nh\mathrm{...}\left(3\right)$
Solving equation (1) and (3) for  r we get
 $r=\frac{4\pi {\epsilon }_0{n}^2{h}^2}{\mu e^2}\mathrm{...}\left(4\right)$
Electrical potential energy of the system is  $U=\frac{e^2}{4\pi {\epsilon }_{0}r}$
and kinetic energy is $K=\frac{1}{2}I{\omega }^2=\frac{1}{2}\mu r^2{\omega }^2$ 
But from equation (1)
${\omega }^2=\frac{e^2}{4\pi {\epsilon }_0\mu r^2}\therefore K=\frac{e^2}{8\pi {\epsilon }_{0}r}$