02 Linear Attenuation Coefficient [PDF]

Zitiervorschau

Experiment 2 Linear Attenuation Coefficient Attenuation of Gamma rays by matter

Objective: 1- Verification of the attenuation equation of Gamma radiation. 2- Determining for gamma rays produced by a certain isotope: a. The mass attenuation coefficient (μm). b. The linear attenuation coefficient (μ) in lead (Pb) and Aluminum (Al). c. The half value layer thickness (X1/2) of the absorbing materials Pb and Al.

Introduction: The experimental set-up is illustrated in the figure below. We refer to the intensity of the radiation which strikes the absorber as the incident intensity, I0, and the intensity of the radiation which gets through the absorber as the transmitted intensity, I. Notice also that the thickness of the absorber is denoted by x.

The transmitted gamma-rays will in the main be those which pass through without any interactions at all. We can therefore expect to find that the transmitted intensity will be less than the incident intensity, that is

1

1.Effect of Atomic Number: Let us start exploring the magnitude of ∆I by placing different absorbers in turn in the radiation beam. What we would find is that the magnitude of ∆I is highly dependent on the atomic number of the absorbing material. For example we would find that ∆I would be quite low in the case of an absorber made from carbon (Z=6) and very large in the case of lead (Z=82). We can gain an appreciation of why this is so from the following figure:

In the case of the low atomic number absorber however the individual atoms are smaller and hence the chances of interactions are reduced. In other words the radiation has a greater probability of being transmitted through the absorber and the attenuation is consequently lower than in the high atomic number case. 2.Effect of Density A second approach to exploring the magnitude of ∆I is to see what happens when we change the density of the absorber. We can see from the following figure that a low density absorber will give rise to less attenuation than a high density absorber since the chances of an interaction between the radiation and the atoms of the absorber are relatively lower. In addition, the density determines the transmission coefficient as it relates to the sample, since the lower the density, the higher the transmission coefficient due to the porous nature of the material.

2

3.Effect of Thickness A third factor which we could vary is the thickness of the absorber. As you should be able to predict at this stage the thicker the absorber the greater the attenuation 4.Effect of Gamma-Ray Energy The greater the energy of the gamma-rays the less the attenuation.

Theory: When Gamma radiation passes through matter, it undergoes attenuation primarily by Compton, photoelectric and pair production interactions. The intensity of the radiation is thus decreased as a function of thickness of the absorbing material. The mathematical expression for intensity ( I ) is given by the following expression: 𝐼 = 𝐼𝑜 𝑒 −𝜇𝑥

(1)

where, Io is the original intensity of the beam. I is the intensity transmitted through an absorber thickness X. μ is the linear attenuation coefficient for the absorbing material. (Note that the two terms attenuation and absorption are often used interchangeably even though they differ slightly. Attenuation is the result of both absorption and scattering of the photons in the material). If we rearrange eq.(1) and take the natural logarithm of both sides, the expression becomes, 𝐼

ln ( ) = −𝜇𝑥 𝐼 𝑜

(2)

The half value layer (HVL) of the absorbing material is defined as that thickness X1/2 which will decrease the initial intensity by half. That is, IHVL=Io/2. If we substitute this into eq.(2) we get: ln(2) = 𝜇𝑥1𝑙2 3

(3)

Rearranging eq.(3) we get: 𝑥1𝑙2 =

ln(2)

(4)

𝜇

Keep in mind that both the attenuation coefficient and the HVL are functions of the energy of the gamma photons. Another useful concept is the mass attenuation coefficient (µm) defined by: μm =

μ

(5)

ρ

Where  is the density of the absorber material. This concept is useful because attenuation is found to be dependent on the density of the absorber material (more atoms and heavier atoms in the way of the photons mean more interactions per unit length of the material). Therefore, both the linear attenuation coefficient and the HVL will have different values for the same material depending on its phase (solid, liquid or vapor) and on its temperature. By dividing on density however, we get a quantity that is independent of the density and hence is the same for the material regardless of its phase and temperature. This quantity is the mass attenuation coefficient. Moreover, μm is approximately independent of the type of material. We can also define the equivalent thickness (or mass thickness𝑥𝑚 ) to be 𝑥𝑒𝑞 = 𝜌𝑥 Writing equation (1) in terms of these two quantities we get 𝐼 = 𝐼𝑜 𝑒

−(𝜇𝑚 𝜌)(

𝑥𝑒𝑞 ) 𝜌

=𝐼𝑜 𝑒 −𝜇𝑚 𝑥𝑒𝑞

(6)

Which is the same as eq (1) and hence 𝜇𝑚 𝑎𝑛𝑑 𝑥𝑒𝑞 can be used instead of µ and x to find all results and they satisfy: (𝑥𝑚 )1/2 =

ln(2) 𝜇𝑚

(7)

Note that taking natural logarithm of both sides of eq(6) gives: 𝐼

ln( ) = −𝜇𝑚 𝑥𝑒𝑞 ⟹ ln(𝐼) = −𝜇𝑚 𝑥𝑒𝑞 + ln(𝐼𝑜 ) 𝐼𝑜

(8)

From which we can see that if we draw a graph of ln(𝐼) vs 𝑥𝑒𝑞 , then the slope of that graph would be −𝜇𝑚 . 4

Apparatus: Source of gamma radiation. Sheets of different absorbing materials (Aluminum and Lead) and different thicknesses . Nuclear station ST150.

Procedure: 1- Connect the plugs of the electric mains. 2- Set the timer to 60s and the voltage to the operating voltage found in first lab. 3- Record the count rate per one minute for the back ground (IB.G) 3 times and take the average. 4- Put the source in front of the GM tube on the 3rd shelf from the top. 5- Record the count rate ( Io ) 3 times and take average. 6- Place Al sheet between the source and the GM tube. 7- Record the count rate 3 times (I1 , I2 and I3) and then find Iavg.. 8- Repeat steps 6 and 7 for the rest of the aluminum sheets with increasing the thickness of the absorbing material. Also take all possible combinations of 2 Al sheets between the source and the GM tube. 9- Repeat steps 6,7 and 8 for the Pb sheets (4 sheets plus 4 combinations of two sheets to get 8 data points). For each material do the following steps: 10- Plot a graph between ln(I ) and thickness 𝑥𝑒𝑞 , if the relation is a straight line, then the attenuation law is verified. 11- Find the slope from the graph. 12- Find the negative of the slope from the graph, this is equal to the mass attenuation coefficient. 13- Calculate the linear attenuation coefficient using eq (5). Note that: Al= 2.7 g/cm3 , pb=11.34 g/cm3 14- Calculate the HVL thickness by using eq.(4).

5

Data Sheet Experiment 3: Attenuation of Gamma rays by matter Source description Element

Activity (A0) Half life (t1/2) ( ……. ) ( ……)

Date of Calibration

Data: IB.G = (……..+.……+…….)/3= ……….. (….…). Io= (……..+.……+…….)/3= ……….. (….…).

material

Sheet numbers

𝑥𝑒𝑞 (……...)

I1 (….....)

Iav I2 I3 I=Iav – IB.G (….....) (……..) (….....) (.…….)

Al

Pb

6

Ln (I)

Calculations and Results:  Part 1: Aluminum (Al)  Slope of graph of ln(I) vs 𝑥𝑒𝑞 = ………… (

)

 mass attenuation coefficient μm = ……….. (

)

 Linear attenuation coefficient 𝜇 = ……….. = ……….. = ………….(  HVL thickness = 𝑥 =

=

= …………(

)

)

 Part 2: Lead (Pb)  Slope of graph of ln(I) vs 𝑥𝑒𝑞 = ………… (  mass attenuation coefficient μm = ……….. (

) )

 Linear attenuation coefficient 𝜇 = ………..= ……….. = ………….(  HVL thickness = 𝑥 =

=

= …………(

) )

Post lab questions:  Find the thickness of Aluminum that would reduce the intensity of the given gamma radiation to one tenth of its original intensity. Write full equations.

 Find the thickness of Lead that would reduce the intensity of the given gamma radiation to one tenth of its original intensity. Write full equations.

 Compare. Which material is better to use for shielding?

7