Chemistry - Why is absorbance spectrum continuous and not quantized?

Solution 1:

The image shows the absorption and emission spektra of molecules. For isolated atoms, you would indeed observe sharp lines. Think in the Fraunhofer lines.

UV/VIS absorption of molecules

In the case of molecules, absorption occurs from the vibrationally relaxed electronic ground state ($S_0$) to various vibration levels of the electronically excited $S_1$ state. This results in a broad absorption band instead of a sharp line.

The manifold of $S_1$ with different vibrational levels does not live forever. Within the shortest time, relaxation to the vibrational ground state of $S_1$ occurs. The electron distribution of the excited state is typically significantly different from the $S_0$ state. This results in a completely changed reactivity.

Radiationless deactivation

If you don't observe emission from an $S_1$ or $T_1$ state, the excited state is deactivated by internal conversion, i.e. by vibrational exchange with the bulk (solvent).

Solution 2:

The pictures you show are misleading as they refer only to molecules in the condensed phase. Isolated molecules in the vapour phase do have discrete (quantised) electronic, vibrational and rotational levels that can easily be measured, and have been so for at least 50 years.

In the condensed phase the spectral lines are inhomogeneously broadened, this means that what you observe is a mixture of many superimposed transitions making a broad structureless feature. This occurs because molecules suffer random collisions/interactions with solvent molecules that shifts energy levels up and down just a little. There are so many different ways this happens that the individual (narrow) spectral line become overlapped and the measured spectrum becomes broad.

We know this because if you freeze a solution close to zero K and illuminate it with a very narrow wavelength laser it is possible to 'burn away' some of the molecules and leave a hole in the spectrum. This is seen as a narrow dip in the spectrum. The hole itself has a width due to the intrinsic lifetime of (say) the excited vibrational transition.

To answer your other points, fluorescence is emitted and either escapes completely into space or is absorbed by some other molecule. The molecule after fluorescing is often left with some vibrational energy in its ground state, this equilibrates with its surroundings as heat. If a triplet is formed, then phosphorescence can be emitted and so on as for fluorescence. If internal conversion (S1 to S0 radiationless transition), all the electronic energy appears as vibrational energy ( a very 'hot' ground state) and then equilibrates with surroundings.