The ratio of the areas of the electron orbits for the second excited state to the first excited state for the hydrogen atom is

According to Bohr's model, the highest kinetic energy is associated with the electron in the

In any Bohr orbit of the hydrogen atom,the ratio of kinetic energy to potential energy of the electron is

When an electron jumps from the orbit $n=2$ to $n=4$,the wavelength of the radiation absorbed will be ($R$ is Rydberg's constant).

When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state,the de-Broglie wavelength associated with it

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 $R_0 = 1 \,\mathring{A}$.