In the Bohr model of a hydrogen atom,the centripetal force is furnished by the Coulomb attraction between the proton and the electron. If ${a_0}$ is the radius of the ground state orbit,$m$ is the mass,$e$ is the charge on the electron,and ${\varepsilon _0}$ is the vacuum permittivity,the speed of the electron is:

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

The de-Broglie wavelength of an electron in the $n^{th}$ Bohr orbit is $\lambda_n$ and the angular momentum is $J_n$,then:

The difference between the radii of $n^{\text{th}}$ and $(n+1)^{\text{th}}$ orbits of a hydrogen atom is equal to the radius of the $(n-1)^{\text{th}}$ orbit of hydrogen. The angular momentum of the electron in the $n^{\text{th}}$ orbit is $.........$ ($h$ is Planck's constant).

The magnetic moment of an electron due to its orbital motion is proportional to $(n =$ principal quantum number$)$

The Bohr model for the $H$-atom relies on Coulomb's law of electrostatics. Coulomb's law has not been directly verified for very short distances of the order of $\mathring{A}$. Suppose Coulomb's law between two opposite charges $+q_1$ and $-q_2$ is modified to $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{r^2} \right)$ for $r \ge R_0$ and $|\vec{F}| = \frac{q_1 q_2}{4\pi \epsilon_0} \left( \frac{1}{R_0^{2-\epsilon} r^{\epsilon}} \right)$ for $r < R_0$. Calculate the ground state energy of an $H$-atom,given $\epsilon = 0.1$ and $R_0 = 1 \,\mathring{A}$.