# Does broader peak mean stronger absorber in UV/VIS?

I want to ask a question about peaks in a UV/VIS spectrum.

My lecturer in our introductory course mentioned the following:

Two molecules may have the same molar extinction coefficient, but if I have two molecules, A and B, and B has a peak that is broader than molecule A, it will be a stronger absorber.

He failed to qualify this further or provide any reasoning.

My thought was that if it is a broader peak, then molecule B will absorb a wider range of wavelengths in a UV/Vis spectrum, but I couldn't see what he meant by a "stronger" absorber.

Does he mean that molecule B absorbs more wavelengths, so it is "stronger" or is there another qualification for using the word "stronger" to describe B?

• Tell him you are not paying tuition fees to be taught dubious facts which you then have to learn by heart. – Karl Nov 16 '19 at 21:09

The classical concept of oscillator strength $$f$$ is useful here, but it should be used only in a qualitative way. This is defined as $$f=a\int\epsilon_\nu d\nu$$ where $$a$$ is set of units with value $$4.3 10^{-9}$$ if the extinction coefficient is in units dm$$^3$$/mol/cm and $$\nu$$ is in wavenumbers. The maximum value of $$f$$ is unity and is close to this for an allowed transition in a dye molecule, such as rhodamine, but can be as small as $$10^{-9}$$ for a forbidden transition.
In your case if two molecules have similar $$\epsilon$$ at some wavelength (or frequency) but one has a narrow transition compared to the other, then the one with the narrow transition will have a smaller integral and hence the smaller oscillator strength. In this sense only is the transition 'weaker', i.e. it is not the probability of absorbing at a given wavelength that is being considered but that over all wavelengths, so it really depends on how you want to define a 'strong' vs a 'weak' transition. In QM the transition moment integral is used instead of oscillator strength and gives a proper description of absorption.