The energy of a hydrogen atom in its ground state is $-13.6 \,eV$. The energy of the level corresponding to $n=7$ is ...... $eV$.

The energy of a hydrogen atom in the $n^{th}$ orbit is $E_n$. What will be the energy in the $n^{th}$ orbit of a singly ionized helium atom?

The energy required to knock out the electron in the third orbit of a hydrogen atom is equal to

When a hydrogen atom is raised from the ground state to an excited state,

When an electron in a hydrogen atom jumps from the third orbit to the second orbit,it emits a photon of wavelength $\lambda$. When it jumps from the fourth orbit to the third orbit,the wavelength emitted by the photon will be

The ratio of energies of photons produced due to transition of an electron in a hydrogen atom from the second energy level to the first energy level and from the fifth energy level to the second energy level is: