The ratio of the kinetic energies of the electrons in the third and fourth excited states of the hydrogen atom is

Assertion : Between any two given energy levels,the number of absorption transitions is always less than the number of emission transitions.
Reason : Absorption transitions start from the lowest energy level only and may end at any higher energy level. But emission transitions may start from any higher energy level and end at any energy level below it.

An electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases is

$A$ hydrogen atom absorbs radiation of wavelength $975 \, \text{\AA}$ to transition from the ground state to an excited state. To which energy level will the atom transition?

If the electron in a hydrogen atom were in the energy level with $n = 3$,how much energy in joule would be required to ionise the atom? (Ionisation energy of $H$-atom is $2.18 \times 10^{-18} \ J$):

$A$ stationary hydrogen atom has an electron that transitions from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission will be: ($m$ is the mass of the atom,$R$ is Rydberg constant,and $h$ is Planck's constant).