The kinetic energy of the electron in an orbit of radius $r$ in a hydrogen atom is proportional to ($e=$ electronic charge).

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:

Assuming the atom is in the ground state, the expression for the magnetic field at the nucleus in a hydrogen atom due to the circular motion of the electron is [$\mu_0 =$ permeability of free space, $m =$ mass of electron, $\epsilon_0 =$ permittivity of free space, $h =$ Planck's constant].

The radius of a certain orbit of a hydrogen atom is $8.48 \mathring{A}$. If the energy of the electron in this orbit is $E/x$,then $x = . . . .$
(Given $a_0 = 0.529 \mathring{A}$,$E =$ energy of the electron in the ground state)

When a hydrogen atom absorbs a photon of wavelength $60 \ nm$,the atom undergoes photo-ionization. What will be the maximum kinetic energy of the emitted electron in $eV$?

If $\lambda_1$ and $\lambda_2$ denote the wavelengths of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom,then $\lambda_1/\lambda_2$ is equal to