If an electron is revolving around the hydrogen nucleus at a distance of $0.1 \,nm$, what should be its speed?

Suppose an electron is attracted towards the origin by a force $F = \frac{k}{r}$,where $k$ is a constant and $r$ is the distance of the electron from the origin. By applying the Bohr model to this system,the radius of the $n^{th}$ orbital of the electron is found to be $r_n$ and the kinetic energy of the electron is $K_n$. Then which of the following is true?

The energy of an electron in the ground state of a hydrogen atom is $E$. What is the energy of an electron in the $2^{nd}$ excited state of $Li^{++}$?

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} \times 10^5 \text{ m s}^{-1}$. The radius of the orbit is $x \times 10^{-9} \text{ m}$. The value of $x$ is : [Take the mass of electron to be $9 \times 10^{-31} \text{ kg}$, charge of electron = $-1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$]

If $m$ is the mass of an electron,$v$ is its velocity,and $r$ is the radius of a stationary circular orbit around a nucleus with charge $Ze$,then from Bohr's postulate,the kinetic energy $K = \frac{1}{2}mv^2$ of the electron in the $C.G.S.$ system is equal to:

If the radius of the first Bohr orbit is $r$, then the radius of the second Bohr orbit will be