In $(i)$ a hydrogen atom and $(ii)$ a singly ionized helium atom,an electron makes a transition from the same excited state $n$ to the ground state. The ratio of the wavelengths of the emitted photons in the two cases will be:

Using Bohr's atomic model,derive an equation for the radius of the $n^{th}$ orbit of an electron.

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).

Monochromatic radiation is incident on a hydrogen $(H)$ sample which is in the ground state. If the hydrogen atoms emit radiation of $10$ different wavelengths after absorbing the incident radiation, then the wavelength of the incident radiation is (Let $hc = 1242 \text{ eV-nm}$) (in $\text{ nm}$)

An electron in the hydrogen atom excites from the $2^{nd}$ orbit to the $4^{th}$ orbit. The change in angular momentum of the electron is (Planck's constant $h = 6.64 \times 10^{-34} \ J \cdot s$).

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