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).

Consider the $3^{rd}$ orbit of $He^{+}$ (Helium) using a non-relativistic approach. The speed of the electron in this orbit will be (given $K = 9 \times 10^9 \; N \cdot m^2/C^2$,$Z = 2$,and $h = 6.6 \times 10^{-34} \; J \cdot s$).

What is the minimum energy that must be given to a $H$ atom in the ground state so that it can emit an $H_{\gamma}$ line in the Balmer series? If the angular momentum of the system is conserved,what would be the angular momentum of such an $H_{\gamma}$ photon?

The gravitational attraction between an electron and a proton in a hydrogen atom is weaker than the Coulomb attraction by a factor of about $10^{-40}$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction.

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

When an electron is excited from its $4^{\text{th}}$ orbit to its $5^{\text{th}}$ stationary orbit,the change in the angular momentum of the electron is approximately (Planck's constant $= h = 6.63 \times 10^{-34} \text{ J s}$)