Question

1. in the mercury emission spectrum, three intense emission bands can be observed. in the Bohr model of the atom, each of these bands corresponds to a transition from a particular quantum orbital (energy level)n= 1,2,3 .....
for mercury, what color band corresponds to the largest energy difference between levels and which is the smallest? why is that?

2. is it possible for an atom to absorb a photon of red light and remit a photon of violet light? why?

Answer 1

1. For mercury, the band of color VIOLET represents the largest energy difference. The ORANGE colored band represents the smallest energy difference.

This is due to the fact that, the wavelength of VIOLET colored band is the least and for the ORANGE band it is the highest. This is because, energy difference between two Bohr states is inversely proportional to wavelength of the emitted light.

ΔΕ =
       where $\Delta E$ is the energy difference between two states, h is the Plank's constant, c is the velocity of light in vacuum and $\lambda$ is the wavelength of the emitted band.

2. It is not possible for an atom to absorb a photon of red light and remit a photon of violet light. The RED light photon is energetically inferior to a VIOLET light photon.As it is not possible for an atom, or any system (in their ground states) to absorb lower energy and emit higher energy thereafter, an atom cannot give off VIOLET light photons after absorbing RED light photons.