Question

Gamma-rays (Y-rays) are highly energetic electromagnetic radiation that are emitted when the nuclei undergo transitions in the energy levels. The energy of the y-rays is upward of 100 keV. Due to the high energy, the rays are not appreciably absorbed by even several centimeters of materials such as concrete or wood. However, dense metals such lead, are effective Y-ray absorbers. The amount y-rays that are absorbed depends on the material and the energy of the gamma-rays, the higher the energy, the more penetrating the rays, i.e., less is absorbed. Gamma-rays interact with materials by three primary methods, (1) Compton scattering, (2) photoelectric effect and (3) pair production. The last method is opresent only when the energy of the gamma-rays equal or greater than 2moc where mo is the rest mass of the electron and c is the speed of light. (Learn more about the absorption of gamma rays.) The amount of y-rays that are absorbed by a material is measured using a setup shown in the schematic in Figure 1. For each thickness of a material that is placed between the y-ray source and the detector, the number of counts (intensity) is measured for a set length of time. (For the actual lab, this time is set at 2 minutes.) The absorption of the y-rays follows a statistical process. (The lab manual has a more detail description of the process.) This results is the intensity of the form where to is the incident intensity, that is, I = To when x = 0. I is the intensity left after the radiation traverses a thickness x of material, u is the absorption coefficient of the material and e = 2.718... is the base of the system of natural logarithms. Figure 1 - Schematic of the experimental setup. Note: There are three exercises for this pre-lab. As you complete one exercise or have exhausted all your tries, the next exercise will be displayed since it may depend on the results of the earlier exercise(s). In addition, this significantly reduces the time to process each submission. When you have exhausted all your tries, the computer's answer will be displayed. Exercise 1 - Absorption by Lead The table below lists the thicknesses of lead and the intensities. Our object is to calculate the absorption coefficient, and the thickness when the intensity is half Io. Since the intensity has an exponential dependence on thickness (Equation 1), a linear plot of intensity vs thickness is not very useful. However, by taking the natural log on both sides of Equation 1, we have in () = n(A). This is an equation for a straight line, the slope of which is u. Thickness (cm) Intensity (counts) 0 6555 0.3 4929 0.6 3491 0. 9 2548 1.2 1897 1.5 1384 2 811 .) Make a plot of In (I) vs thickness and find the slope. Then enter the value below for u. (Note: the slope gives you - and it has unit of cm (Lead) - 1.05 cm-1 You are correct. Your receipt no. is 162-6739. Previous Tries Shown below are two plots, the first is a linear plot while the second is a loge-linear plot. The linear plot clearly shows an exponential dependence of the intensity with thickness. From the loge-linear plot, the absorption coefficient, (= minus of the slope) can easily be determined. Intensity vs thickness plot A plot of In (I) vs on a linear scale. The plot thickness x gives a shows an exponential straight line where the decay. slope is -u. A useful quantity to specify for an absorbing material in addition to its absorption coefficient is its half-value thickness, Xy. This is the thickness that absorbs half of the incident y-rays. The half-value thickness may be found in two ways. Either the value for X at which the count rate drops to one-half its initial value may be read directly from the graph (left plot above) or X1, may be determined from u from the relation, 4 = 42 Find the value of X1/ for lead and enter the value below. X1 (Lead) = cm Submit Answer Tries 0/10

Answer 1

olna NS 0,693 Op 0.693 1,05 = 0.66 cm (od0,669 am cm