Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Material Type: Lab; Class: Advanced Laboratory; Subject: PHYSICS; University: University of Wisconsin - Madison; Term: Spring 2009;
Typology: Lab Reports
1 / 19
This page cannot be seen from the preview
Don't miss anything!
(revised 3/19/09)
Compton Scattering
Advanced Laboratory, Physics 407 University of Wisconsin Madison, Wisconsin 53706
Abstract
The Compton scattering of the 662 keV gamma rays from the decay of Cs^137 is measured using a Sodium Iodide detector. The scattered energy and the differential cross section are both measured as a function of scattering angle, and the results are compared to the full relativistic quantum theory of radiation.
Theory
The Compton effect is the elastic scattering of photons from electrons. As a reaction, the process is:
γ + e−^ → γ + e−^.
Since this is a two body elastic scattering process, the angle of the scat- tered photon is completely correlated with the energy of the scattered photon by energy and momentum conservation. This relation is usually written as:
∆λ = λ′^ − λ =
h mc
(1 − cos θ) (1)
where λ and λ′^ are the wavelengths of the incident and scattered photon respectively, and θ is the photon scattering angle. The energy of a photon is related to its frequency and wavelength as:
E = hν = hc/λ (2)
where c is the velocity of light. Combining Eq. (1) and (2) the energy of the scattered photon is:
1 + (^) mcE 2 (1 − cos θ)
The kinetic energy of the recoil electron is:
Te = E − E′^ = E
γ(1 − cos θ) 1 + γ(1 − cos θ)
where γ = hν/mc^2. The quantity h/mc in Eq. (1) is called the Compton wavelength and has the value: h/mc = 2. 426 × 10 −^10 cm = .02426 ˚A.
For low energy photons with λ .02 ˚A, the Compton shift is very small, whereas for high energy photons with λ 0 .02 ˚A, the wavelength of the scattered radiation is always of the order of 0.02 ˚A, the Compton wavelength. As an example, in this experiment gamma rays from Cs^137 are scattered from an aluminum target; since E = 0.662 MeV, we have γ = 1.29 so that gamma rays scattered at θ = 180◦^ will have E′^ = E/ 3 .6, which is less
The problem was solved by Klein and Nishina in 1928 giving the correct quantum-mechanical calculation for Compton scattering, the so called Klein- Nishina formula:
dσ dΩ
= r^20
( 1 + cos^2 θ 2
) 1 (1 + γ(1 − cos θ))^2
×
[ γ^2 (1 − cos θ)^2 (1 + cos^2 θ)(1 + γ(1 − cos θ))
]
. (7)
This result is for the cross section averaged over all incoming photon polar- izations. By integrating Eq. (7) over all angles, the total cross section can be obtained. While the expression for the total cross section is a lengthy formula, two asymptotic expressions for the total cross section σc in the low energy and high energy case are more simple.
Low energies (γ 1) σc = σT
{ 1 − 2 γ +
γ^2 + · · ·
}
and High energies (γ 1) σc =
σT
γ
( ln 2γ +
)
. (8)
Note that Eq. (5) for the Thompson cross section gives a symmetric angular distribution of scattered photons (i.e., the angular distribution is symmetric about 90◦). The Klein-Nishina formula (Eq. (7)) on the other hand predicts a strongly forward peaked cross section as γ increases. The following table and graph show the Klein-Nishina cross section as a function of photon scattering angle. The table is for the .662 MeV gamma ray energy of Cs^137 , while Fig. 2 shows the angular distribution for a range of incident photon energies. Radiation Units Milliroentgen per hour (mR/h) are units of radiation exposure. Expo- sure indicates the production of ions in a material by radiation, and it is
(Degrees) cos θ Classical E (keV) Relativistic E (keV) Klein-Nishina
- 0 1.00000 662.00 662.00 7. (10−^30 m^2 ) - 5 .99619 658.74 658.75 7. The effects of radiation on biological systems depend on the type of ra- diation and its energy. The relative biological effectiveness (RBE) or quality factor (QF) of a particular radiation is defined by comparing its effects to those of a standard kind of radiation, which is usually taken to be 200-keV x rays. The RBE or QF is the ratio of the dose in rads of a particular kind of radiation to a 1-rad dose of 200-keV x rays, where the particular radiation produces the same biological effect as the x rays. Note that RBE or QF is dimensionless.
RBE or QF:
Number of rads of a particular kind of radiation 1 rad of 200-keV x rays
In animal tissue the RBE is about 1.0 for x rays, γ rays, and β rays, and it ranges from about 2 to 20 for protons, neutrons, and α particles. The rem is defined as
rem ≡ dose in rads × RBE.
For animal tissue a 1-rad dose of γ rays is equivalent to an exposure of 1 R, and the RBE is about 1 for γ rays in animal tissue; therefore, the dose in rems and the exposure in roentgens are equivalent. Radiation standards adopted by the United States Government are the following:
Shielding Calculations The calculations consist of two parts. First we calculate the attenuation of gammas which is desirable. Then we calculate the thickness of lead to give this attenuation.
x = 0.43 R/hr (with no shielding)
B This corresponds to a dose equivalent D for tissue of .415 rads/hr. C Converting this to the Dose Equivalent in rems.
DE = D × QF
where QF=1 for gamma rays. Hence we have
DE = 0.415 rems/hr (if no shielding)
D Assume that a student may spend 16 hours/month standing at this short distance (0.3 meters) from the source. The Dose equiv- alent/year with this occupancy factor is then
79.7 rem/year (if no shielding)
E Radiation standards adopted by the U.S. (1990) for worldwide population: 5 rem total up to age 30 (0.17 rem/yr), in addition to natural background radiation, which is about the same inten- sity. We will assume that 1 × 10 −^3 rem/yr is satisfactory for the shielding of our Cs^137 source. We have therefore designed the lead shielding to give an attenuation factor of:
f =
1 × 10 −^3 rem/yr 79 .7 rem/yr
A shielding calculation (not shown here) for 662 keV gamma rays gives the result that 10.5 cm of lead will accomplish the above attenuation factor.
Apparatus
Gamma Source The gamma source is Cs^137. The present source had a nominal strength of 100 millicuries when it was purchased in 1975, although the actual strength
Outer Plug with 8-32 Threaded End Outer CapsuleInner Plug Inner Capsule Activity
3/8’’ Hex 1–1/2’’ TYPE 193 GAMMA SOURCE CAPSULE ALL FUSION WELDED STAINLESS STEEL CONSTRUCTION Isotope Products Inc. Model HEG-137-
Figure 4: Cs^137 Source Details
Compton Effect. The most important angle of scattering is in the forward ± 45 ◦^ cone because
(a) The differential cross section is larger in the forward direction, and
(b) The minimum energy is lost by the gamma when it is scattered in the forward direction.
As we require a jet of well collimated 662 KeV gammas with as little con- tamination as possible of gammas with lower energies, the container barrel has been designed to greatly reduce the chance of gammas escaping after a forward scatter (θ < 90 ◦). If a 662 KeV photon is scattered through more than 45◦^ then it is likely to be absorbed by a small thickness of lead.
N = N 0 e−x/λ
Angle of Scattering Energy λ in Lead 0 ◦^ 662 keV 0.88 cm 45 ◦^ 479 keV 0.57 cm 90 ◦^ 288 keV 0.22 cm 135 ◦^ 206 keV 0.11 cm 180 ◦^ 184 keV 0.09 cm
The gammas which do not pass out through the aperture are incident upon the surfaces A, B, C, D. (See Fig. 5.)
Lead
Lead
A
B
C^ D (^) E
F
O^ source
Figure 5: Source Collimation
A and B A few gammas will Compton Scatter to give 184 MeV gammas which scatter to the right. However, the 662 gammas will have penetrated about 0.88 cm into the lead and so only a few of the 184 MeV gammas will escape from inside the AB surfaces. Some gammas will interact via the photoelectric effect. Both will cause 75 keV X-rays (Lead K shell) but again only a few X-rays will escape from inside the AB surfaces. C The aperture is designed so that scattered gammas and X-rays cannot escape directly from C. D Again scattered gammas and X-rays cannot escape directly. Although the Compton scatters of 90◦^ to 45◦^ may have enough energy for a second Compton scatter on the opposite surface E, the distance they must travel from the interaction point in D to leave D is greater than the penetration into D. Note that a few gammas will Compton Scatter near the DE corner and will escape into the main beam. Obviously if we had a ring of dense HIGH Z material such as uranium, at the corner, this background could be reduced. (See Fig. 6)
D E
lead
Gamma Gamma
Figure 6: Source Collimator Scattering
E This is a conical surface designed so that it cannot be directly illumi- nated by gammas from the source.
Angle of scintillator subtended at the scatterer ± 8 ◦ Minimum angle of scattering which is outside the jet: 15◦. Tapered Plug The plug is intended to block the beam. The taper is made from brass and so must be longer than if it had been made from lead. We except a few scattered gammas to sneak along the taper since the fit cannot be perfect. For this reason an extra 3.8 cm of steel is used to block those escaping along the taper.
Detector (Harshaw Chemical Type 858 Serial 6V230 with voltage divider) The gamma detector is a NaI crystal scintillation cylinder with a diameter of 2 inches and a thickness of 2 inches. The crystal is hermetically sealed and in good optical contact with the photocathode of an RCA 6342A phototube with 10 stages. The Harshaw Quality Assurance Report states that the detector has a resolution when measuring 662 keV gammas of
full width at 1/2 maximum counting rate pulse height
Read the description of the experiment on Interactions of Gammas for further information. The photomultiplier requires a positive high voltage via the white cable and High Voltage BNC connector. The dynode potentials are controlled by the built in voltage divider with a total resistance of 6.2 Mohms. The anode is at the positive high voltage and so a 1 nf capacitor is used to pick off the 1 μsec 100 millivolt negative signals. The maximum HV rating is +1500V but the experiment needs only ∼ 1100–1200 V. The NaI-PM assembly is mounted inside a tapered lead cylinder. The lead weighs about 23 kg and is intended to serve as a collimator so that only gamma rays from the target region are detected. See Fig. 8.
γ
Pb NaI^ PM
Figure 8: Detector Geometry
The NaI crystal diameter is 2 inches but only 1 3/4 inches are exposed to the gammas to give the greatest probability of full absorption of each gamma entering the NaI. The lead cylinder is mounted on rollers and rotates about a center which can support a scatterer.
Scatterer The scatterer is an aluminum disc (1.27 cm thick, 3 cm diameter). The Compton effect depends only on the electrons but does assume that the electrons are loosely held. For this reason, the Z of the scatterer should not be too high. The K shell electrons in aluminum have a level of −1560 eV and the L shell electrons have levels of −118, −74 and −73 eV. The scatterer was chosen as a disc so that the probability of multiple scattering may be minimized. The scatterer is mounted on rigid foam plastic so that the support will scatter few gammas.
Spectrum Techniques UCS- The UCS-30 is a unit that contains a preamplifier, high voltage supply, and Pulse Height Analyzer (PHA). The UCS-30 software controls all the settings. The nominal settings are +1174 V for the HV, Coarse Gain = 32, and Fine Gain = 1.75. We normally use 512 channels for the PHA. The amplifier gain is set to match the 0-8 V range of the PHA and is set so that the Cs^137 peak falls at about 90% of the maximum PHA channel.
Procedure
of 15◦, going out to 135◦^ The scatterer should be rotated so as to approximately bisect the angle between the source and the detector. (Why? The gammas are not reflected like light—why not?) Use your knowledge of counting statistics to decide how long to count the spectra. At each angle take a run with and without the Al target. These runs do not have to be for equal times if they are correctly normalized. The PHA software may be used to subtract the appropriately normal- ized background. The software has a “strip background” feature that will automatically subtract a background run from a target run and correctly do the subtraction even if the data and background runs are for different times. Consult the UCS-30 manual for details. You can then place a region of interest (ROI) about the subtracted spectrum and the gross counts in the ROI will be recorded as part of the spec- trum display. The spectra data can be saved on the computer hard drive, and the spectrum display can be sent directly to a printer. For each spectrum, identify the channel number of the peak of the Compton scattering. Use the plot of (2) above to compute a scattered energy E 2 for each angle. The calculation of the differential cross sec- tion will, in addition, require a measurement of the scattering rate at each angle, determined by evaluating the total number of counts in the ROI and nomalizing to the time of the run. The rate must subsequently be corrected for the efficiency of the NaI detector and the photofrac- tion. Why are these corrections different? If there is still a significant background under the subtracted spectrum, it may be more accurate to determine the counts that fall within the FWHM of the detected peak and correct back to the full area. For a Gaussian shaped peak, the counts within the FWHM region correspond approximately to
of the total area. Efficiency and photofraction data for a 2′′diam. × 2 ′′ thick NaI crystal are included as figures 10 and 11.
1 E 2
m 0 c^2
(1 − cos θ).
Plot 1/E 2 from your measurements against 1 − (cos θ).
(a) The agreement of your data and nonrelativistic theory. (b) Would you expect good agreement for small θ? (c) Is the experimental data in the form predicted by relativistic the- ory? (d) From the slope, find the rest energy of an electron.
R(θ) = Iγ × Ne ×
dσ dΩ
× ∆Ωdet
where R(θ) is the scattering rate for a 100% efficient detector, Iγ is the source intensity per unit area at the target, Ne is the number of electrons in the target, and ∆Ωdet is the solid angle subtended by the NaI detector. For the case of our Cs^137 source:
Iγ =
present source intensity (Ci) × 0. 94 × 0. 90 × 3. 7 × 1010 4 π (source to scatterer)^2
The efficiency factors include the efficiency of the NaI at energy E′, N aI (E′) and the photofraction at energy E′, pf (E′). Thus the mea- sured rate G(θ) is related to the true rate R(θ) by:
R(θ) =
G(θ) N aI (E′) × pf (E′) and finally:
dσ dΩ
G(θ) Iγ × Ne × ∆Ωdet × N aI (E′) × pf (E′)
Use Figs. 10 and 11 below for the efficiencies.
(g) Could you distinguish between the energy of a photon which suf- fered a single θ = 90◦^ scatter and the energy of a photon which suffered two θ = 45◦^ scatters?
GAMMA RAY ENERGY – MEV
0.1 0.2 0.4 0.7 1.0 2.0 4.0 7.0 10
20
40
60
80
100
PHOTOFRACTION – PERCENT
8’’ d. x 8’’ h.
8’’ d. x 4’’ h.
4’’ d. x 4’’ h.
2’’ d. x 2’’ h.
Figure 11: Calculated Efficiencies and Photofractions for Various Size Thal- lium Activated Sodium Iodide Crystals, W.F. Miller, John Reynolds, W.J. Snow, Rev. Sci. Instrum. 28 , 717 (1957)