To explain his theory,Bohr used

If the energy in the first excited state in a hydrogen atom is $23.8 \, eV$,then the potential energy of a hydrogen atom in the ground state can be assumed to be ....... $eV$.

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:

Hydrogen $(H)$,deuterium $(D)$,singly ionized helium $(He^+)$ and doubly ionized lithium $(Li^{2+})$ all have one electron around the nucleus. Consider the $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ respectively. Then approximately:

If the total energy of an electron in an orbit is positive,then

The ratio of wavelengths for the transition of electrons from the $2^{nd}$ orbit to the $1^{st}$ orbit of Helium $(He^+)$ and Lithium $(Li^{++})$ is (Atomic number of Helium = $2$,Atomic number of Lithium = $3$).