organic chemistry - Why is absorbance spectrum continuous and not quantized?

1) In text books, the absorption event is usually described by a figure shown below¹: light excites electrons to higher states in atom or molecule. The difference between the ground state and the excited one matches the energy of the absorbed light. But electron states in atoms and molecules should be quantized, so why do we observe absorbance spectra that are smooth and continuous but not spiked and quantized?

2) Where does the energy go when the excited electron returns to its normal state? In fluorescence, the electrons return to lower states and excite light at higher wavelength than the absorbed one. But if I had a molecule that did not exhibit fluorescence, just plain absorption, does the molecule emit light at the same wavelength it absorbed or does the energy turn to heat? I mean, the electron cannot be excited forever.

3) What is the physical phenomenon behind absorption coefficient? I have been thinking that it has something to do with the lifetime of the excited state – shorter lifetimes increase the probability of a photon to be absorbed, resulting in larger coefficients. Am I completely wrong?

¹Image taken from http://www.photobiology.info/Visser-Rolinski.html

Answer

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).