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Cal Poly Pomona PHY 432L - Attenuation of Radiation in Matter

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Attenuation of Radiation in MatterIn this experiment we will examine how radiation decreases in intensity as itpasses through a substance. Since radiation interacts with matter, its intensity willdecrease as it travels through a material. The attenuation properties of radiationwill effect how much shielding is necessary and how much dosage one recieves. Sinceradiation deposits its energy as it travels though matter, the amount of dosage onereceives depends on how the radiation is attenuated. Radiation that attenuates fast,i.e. loses its energy in a short distance, will give a higher dosage (near the surface)than radiation that attenuates slowly, i.e. loses its energy over a long distance in thematerial.First we consider how the intensity of gamma particles decreases as they passthrough a material. We will see that the attenuation properties of gamma particles canbe described fairly simply. The attenuation of β and α particles is more complicated.Attenuation of gamma particlesAs gamma particles pass through matter, they interact with the material and getabsorbed. Suppose gamma particles are incident from the left on a slab, and theslab has a thickness of x. After passing through the slab of material, the gammaparticles emerge on the right. Let the intensity of the gammas incident from the leftbe denoted as I0, the initial intensity. Let the intensity of the gammas that emergeon the right, after passing though the slab, be denoted as I(x), the final intensity.The unit for intensity, I0and I(x), is (number of gamma particles)/area/time. Forunits of intensity we could also use energy/area/time. Either type of unit could beused in our applications, but since our detectors measure counts, we will use (numberof gammas)/time. As the slab gets thicker, x gets larger, and I(x) becomes smaller.As x increases, more radiation is absorbed in the material and less passes throughand emerges on the right side.How does I(x) depend on x? To a very good approximation the number of gammasthat pass through the slab decrease exponentially with thickness:I(x) = I0e−αx(1)This equation is referred to as Lambert’s law, and is applicable for linear attenu-ation. The geometry for linear attenuation is that the material is a rectangular slab,and the gamma’s are incident perpendicular to the slab. The attenuation is in a”line” and is uniform in the plane of the slab. The units of α are (1/distance) since1αx is unitless. In practice x often has units of cm, so α will have units of cm−1. Theparameter α is called the linear attenuation coefficient.The linear attenuation coefficient, α, will depend on the type of material and onthe energy of the gamma radiation, Eγ. When gamma particles travel through matter,any interaction that occurs will be with the electrons in the substance. The higher thedensity of electrons, the more the gamma particles will attenuate. That is, if you wantto shield yourself from gamma radiation, you need to put lots of electrons betweenyourself and the source. Since the electron density is approximately proportionalto the density of the material, the attenuation coefficient α depends strongly on thedensity of the substance. For this reason, it is convenient to define a mass attenuationcoefficient by dividing α by the density ρ of the material. Starting with the equationfor linear attenuation:I(x) = I0e−αx= I0eαρ(ρx)(2)I(x) = I0e−µz(3)where z ≡ ρx, and the mass attenuation coefficient µ is defined as α/ρ. The linearattenuation coefficient is just the mass attenuation coefficient times the density of thematerial:α = µρ (4)The units of µ are (area/mass), and are commonly given in (cm2/g). Since µz isunitless, z is in units of (mass/area) or (g/cm2). The ”thickness” z can be thought ofas the number of grams of ”thickness” per square centimeter area of the slab. Thatis, for one square centimeter of area, z is the number of grams of the substance thatthe gamma’s travel through. The more mass there is, the more electrons there are,and the more attenuation. Materials with a high density will have a large z (lots ofelectrons) for a small thickness, and are good for shielding gammas.In tables of attenuation coefficients, it is the mass attenuation coefficient that isalways listed. This is because the mass attenuation coefficient, µ, has the density de-pendence somewhat factored out. The mass attenuation coefficient mainly dependson energy. For a fixed energy, µ is roughly the same for a large class of materials.When performing an attenuation calculation, one looks up the mass attenuation co-efficient in the tables, multiplies by the density of the substance to obtain α, thenuses the linear attenuation equation.Why does the attenuation follow an exponential formula? Such a simple relation-ship must derive from a simple reason. The derivation of Lambert’s law can obtained23by considering a very thin slice of the slab of thickness ∆x as shown in the figure.Let I(x) be the intensity of the radiation incident on the thin slab from theleft, and let I(x + ∆x) be the intensity of the radiation after passing through thethin slab. From quantum mechanics, the interaction of gamma particles (photons)with electrons is probabilistic. Thus, any particular gamma particle has a certainprobability of being scattered with an electron in the thin slab. Since the numberof electrons the gamma will encounter is proportional to ∆x, then for small enough∆x the probability that a gamma will interact with an electron in the thin slab isproportional to its thickness:P robability for an interaction = α(∆x) (5)As with radioactive decay, probability enters in the description of nature. In radioac-tive decay, the parameter λ was equal to the probability to decay per unit time. Here,the parameter α equals the probability for the gamma to interact per unit length.Now if one has I(x) particles at a distance x, and I(x + ∆x) particles at a distancex + ∆x into the slab, the number of particles lost per area is just the differenceI(x) − I(x + ∆x). However, the number lost per area equals the number incidenttimes the probability for any one gamma to interact with an electron in the slab:I(x) − I(x + ∆x) = (probability for one gamma to interact)I(x) (6)I(x) − I(x + ∆x) = α(∆x)I(x) (7)Dividing both sides by ∆x and multiplying by -1 givesI(x + ∆x) − I(x)∆x= −αI(x) (8)Taking the limit as ∆x goes to zero gives a simple differential

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