Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

So I'm pretty sure I'm approaching this problem in the wrong way and I need some guidance (my first hint is that I think I'm thinking about a quantum mechanical problem too classically)

Suppose there is an isolated molecule in the gas phase with an average cross-sectional area to be exposed to radiation of $A$. (For my specific problem, the molecule is trapped in a superfluid Helium droplet, but I think the calculation should be roughly the same). If the radiation source has a flux $f$ (in units of photons/second/square area/0.1% BW) at energy $E$, what is the probability of the molecule absorbing a photon within a given interaction time $t$ if the absorption probability at a given energy is $P(E)$?

This is pretty easy to calculate if I treat the whole problem classically, i.e. like a ball and a target model. For some reason, though, I get numbers that seem to be way too low if I do this. I know it has to be more complicated than that, since light is a wave also. What am I missing?

I understand that transition probabilities are related to wavefunction overlap, etc. Also, I should note that the radiation in my specific problem is in the hard x-ray region, though I don't think that should change the answer.

Thanks in advance for your help.

share|improve this question
add comment

3 Answers 3

up vote 1 down vote accepted

the set of quantities you offered us to calculate the result is strange, or at least unusual. In particular, there is nothing such as the (dimensionless) absorption probability $P(E)$ for a molecule to absorb a photon.

The absorption probability is given by the cross section which you called $A$. For every process, one has to determine the cross section again. Usually, the cross section is called $\sigma$ instead of $A$.

So the best thing I could do with your numbers and functions would be to imagine that the cross section in your case was given by $\sigma(E) = P(E)\cdot A$. But it is really misleading to factorize the cross section in such ways. In particular, a universal "cross section area $A$" doesn't mean anything. Molecules are not hard balls that have a universal well-defined cross-sectional areas. They're nuclei surrounded by soft wave functions of electrons that reach arbitrarily far but are getting weaker with the distance - in fact, the (one) wave function for $N$ electrons lives in an $3N$-dimensional space rather than the ordinary $3$-dimensional space.

The number of events, in your case absorption events, is simply the product of the integrated luminosity and the cross section $\sigma$. The integrated luminosity is the time integral of luminosity $L$. It is normally written as $L=\rho\times v$ where $\rho$ is the number density of the beam - in particles per unit volume - and $v$ is the velocity - for photons, it's the speed of light $c$. The quantity $L$ happens to be the same thing as your flux $f$. If there are some losses or extra percentages etc., you have to be careful about it.

The quantities such as the cross section are designed in such a way that you are allowed to imagine that the molecule is classical and literally has the cross-sectional area $\sigma$ - that's why they were designed - and you get the right probability. If you do think right, you will indeed find out that only a small percentage of the X-ray photons is absorbed. That's why X-rays are being used to see through human bodies at X-ray pictures.

Best wishes Lubos

share|improve this answer
    
I remember some "oscillator strength" in this context. There is a wikipedia entry for that.Georg –  Georg Jan 17 '11 at 16:37
    
It is quite normal in spectroscopy to define absorption strength of a certain molecule in $cm^2$. Of course the value itself is defined by the transition probability and has no physical meaning but it is very handy for calculations. –  gigacyan Jan 17 '11 at 20:03
add comment

Well my first guess that you shouldn't consider naive "cross-sectional area" of your molecule. It would be better If you calculated the cross section of absorption/scattering of photons on your molecule. It depends on different properties of your molecule and on the wavelength of the light.

But maybe your calculations are more or less correct. You said that you result "too low", but hard x-ray photons are indeed very good at passing through matter.

share|improve this answer
add comment

It is not unusual in spectroscopy to define an absorption cross-section of a molecule in $cm^2$. Although it has no real physical meaning (it can even be larger that the molecule itself) and can be derived from Einstein coefficients, it is what you can directly measure by experiment $$cross-section\;\;\alpha= \frac{total\;energy\;absorbed\;/s}{total\;incident\;intensity\;(energy/s/area)}$$ The relation to the Einstein coefficient if simply $\alpha=\frac{\hbar\omega}{c}B_{21}$

Basically, transition probability is Einstein coefficient times photon density. You problem could be to correctly define a fraction of photons that are resonant with the transition that you study. This would greatly depend on the kind of light source that you use.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.