ABSORPTION CROSS SECTION

I Basic Principles
Collisions are widely used for studying atomic and sub-atomic systems (molecules, atoms, nuclei, nucleons). A beam of particles (e.g., electrons, protons, ions, neutrinos, photons) bombards a sample which contains the target of interest. The target is also an atomic or sub-atomic particle.

In the most accurate experiments, most of the incoming particles, usually referred to as “projectiles”, pass through the sample, while only a few find a target with which to collide. For those that do, the collision is characterized by a quantity known as the cross section. The numerical value of the cross section depends upon the properties of the interacting projectile and target, the strength of the forces they can exert on each other, and the velocity of the projectile (wavelength if the particle is a photon). Sometimes the particle and/or target change identity during the collision. From measurements of the momentum, energy, and charge of all particles before and after the collision, the physicist can obtain information about the fundamental forces by which the projectile and target interact.

When light passes through a sample, photons get absorbed. This means less light leaves the sample than entered. (A “transparent” medium, in which no absorption takes place, is an idealization.) The number of photons that are absorbed depends, at the very least, upon the number of photons that enter, the number of target atoms (“absorbers”) per unit volume in the sample, the length of the sample, and the size of the atoms. Put another way, the probability that a photon is absorbed is proportional to the density and size of the absorbing atoms. More quantitatively, the number dN of photons absorbed, between the points x and x + dx along the path of the beam is given as the product of the number N of photons penetrating to depth x times the number n of absorbers per unit volume and times the absorption cross section σ:

 

dN/dx = –Nnσ

(1)

The cross section will, in turn, depend upon the wavelength of the photons, perhaps upon their polarization, and upon the kind and state of the target atom or molecule.

We can rearrange Eq. (1) and integrate both sides:

 


∫NN0 dN'/N' = ∫t0 nσdx (t is the thickness of the sample.)

 

 

ln(N/N0) = –nσ t

(2)

or,

 

N = N0 exp(–nσ t).

(3)

Eq. (3) is a frequently occurring formula in physics. It applies whenever one particle (or projectile) collides (or interacts) with another particle (the target) present in a target medium. It describes electrons exciting atoms as, for example, in a fluorescent lamp, photons ionizing gas molecules as, for example, by sunlight when it reaches the outer layers of the atmosphere, and slow neutrons initiating nuclear fission in a power reactor, to name just a few applications. In nature there are endless varieties of collisions, and each one has its own unique cross section.

In this experiment you will measure the cross section for absorbing photons from a laser beam as it travels through a solution of copper atoms. In that case, since the number of photons in the beam is proportional to its intensity I, Eq. (3) takes the form

 

I(t) = I0 exp(–αt)

(4)

which is known as Lambert’s Law. The product α = nσ is known as the absorption coefficient. The reciprocal of the absorption coefficient is called the mean-free path, the average distance traveled by a photon before it is absorbed. Whenever the mean-free path is large compared to the size of the sample, the sample is said to be “thin”, and the exponential can be approximated by a linear relation,

 

I(t) = I0 exp(–nσ t) = I0 (1 – nσ t + ··· ) ∼ I0 (1 – nσ t).

 

The transmission T and absorption A are defined as follows:

 

T = I(t)/I0 and A = 1 – T ,

(5)

where t is the sample thickness and I0 is the incident light intensity. When the sample is thin, the absorption is simply

 

A = nσ t (thin sample)

(6)

II Procedure
1) Apparatus (Review Notes on Optical Apparatus).
The setup shown in the diagram below has been prepared in advance, except for the sample.

BE CAREFUL NOT TO LOOK DOWN INTO THE VERTICAL LASER BEAM.

2) Initialization.

    a) Place an empty 150-cm3 beaker so that the vertical laser beam passes upward through the center of the flat window in its base.

    b) Adjust the beaker’s position until the laser beam, as viewed on a piece of paper held above the beaker, forms a small spot, not a line or smudge.

    c) Pour water into the beaker, filling it to the 150-cm3 mark. Adjust the position of the optical fiber tip until a maximum reading is found on the photometer. After this be careful not to move the beaker or the fiber tip. Record the photometer reading as I0. You will need this value to evaluate Eq. (4).

    3) Sample preparation.
    Weigh out five equal-mass samples of copper sulfate. (Your lab instructor will tell how much to put in each sample.)

4) Attenuation measurements.

    (a) Pour one of the copper sulfate samples in the water-filled beaker and stir gently until the crystals are all dissolved. Be careful not to disturb the position or orientation of the beaker.

    (b) Record the intensity of the transmitted light. Call it I1.

    (c) Add a second sample to the beaker, stir, and record the intensity I2 of the transmitted light.

    (d) repeat step (c) until you have used up all the samples, and recorded five values for Ii, i = 1, ..., 5.

    5) Waste disposal.
    Pour the copper sulfate solution into the sink and rinse the beaker with clean water. Wipe any spills and any remaining liquid from the beakers with a paper towel.

III Data Reduction
Plot ln(Ii /I0) versus i, and find the slope of the resulting straight line.

IV Data Analysis
1) Compute the number density

 

n = NAvW/AV

 

where NAv is Avogadro’s number, W is the weight of dissolved copper sulfate used in the first step, A is the atomic weight of the molecule (249.6--see Question #5 below), and V is the solution volume. (Your lab instructor will furnish this value.)

2) Determine the cross section σ from the slope of your plot and the number density n. The cross section you have determined is the cross section for absorbing light of wavelength 632.8 nm by copper ions dissolved in water.

V Questions
1) Was the sample used in this experiment “thin”?
2) What are the units of σ?
3) We have assumed in the analysis that the copper is present in the solution as an atomic ion. How would your calculation of the cross section be effected if the copper were present as a diatomic molecule?
4) Suppose the experiment were done in a vapor with a fixed path length, but with a variable pressure. Explain how you would derive the cross section from a curve of transmitted light intensity versus pressure. Hint: you can assume the vapor obeys the perfect gas law.
5) The copper sulfate crystals used in this experiment are “hydrated”. The technical name is cupric sulfate pentahydrate. Each CuSO4 molecule has five molecules of water attached. Verify the value cited above for the molecular weight.
6) How does your value for the cross section compare with the size of an atom?