experiment 11 - physics

15
Experiment 11 1 The Compton Effect Physics 2150 Experiment 11 University of Colorado Introduction In some situations, electromagnetic waves can act like particles, carrying energy and momentum, which they may impart to other particles with which they interact. According to Planck’s hypothesis, an electromagnetic wave of frequency carries an energy given by: = ℎ (1) where is a positive integer and is Planck’s constant. The minimum amount of energy which can be carried by an electromagnetic wave is thus that carried by one photon. Further, according to the de Broglie relation, if the wavelength of the radiation is , then the photon has a momentum given by = = (2) since 1 = for an electromagnetic wave. Consider a photon which collides with an electron at rest. In Fig. 1, let be the angle at which the photon comes off, and let be the angle at which the electron comes off. Since Figure 1: Collision of a Photon with an Electron Initially at Rest the photon emerging from the collision has a momentum different from that of the incident photon, it will have a different wavelength, which we shall denote by . Applying the relativistic law of conservation of energy to this collision

Upload: others

Post on 08-Jun-2022

20 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Experiment 11 - Physics

Experiment 11

1

The Compton Effect Physics 2150 Experiment 11

University of Colorado Introduction In some situations, electromagnetic waves can act like particles, carrying energy and momentum, which they may impart to other particles with which they interact. According to Planck’s hypothesis, an electromagnetic wave of frequency 𝑓 carries an energy given by:

𝐸 = 𝑛ℎ𝑓 (1) where 𝑛 is a positive integer and ℎ is Planck’s constant. The minimum amount of energy which can be carried by an electromagnetic wave is thus that carried by one photon. Further, according to the de Broglie relation, if the wavelength of the radiation is 𝜆, then the photon has a momentum given by

𝑝 =ℎ

𝜆=

ℎ𝑓

𝑐 (2)

since 1

𝜆=

𝑓

𝑐 for an electromagnetic wave.

Consider a photon which collides with an electron at rest. In Fig. 1, let 𝜃 be the angle

at which the photon comes off, and let 𝜙 be the angle at which the electron comes off. Since

Figure 1: Collision of a Photon with an Electron Initially at Rest the photon emerging from the collision has a momentum different from that of the incident photon, it will have a different wavelength, which we shall denote by 𝜆′. Applying the relativistic law of conservation of energy to this collision

Page 2: Experiment 11 - Physics

Experiment 11

2

ℎ𝑐

𝜆+ 𝑚𝑒𝑐2 =

ℎ𝑐

𝜆′ + 𝑚′𝑐2 (3)

where the mass of the electron after that collision is

𝑚′ =𝑚𝑒

√1− 𝑣2

𝑐2

.

(4) In this equation, 𝑣 is the electron’s velocity in the laboratory reference frame. Applying the laws of conservation of momentum in the plane of the collision gives

𝜆=

𝜆′ cos 𝜃 + 𝑚′𝑣 cos 𝜙, (5)

and

0 =ℎ

𝜆′ sin 𝜃 − 𝑚′𝑣 sin 𝜙. (6)

It follows from these three equations (see any text in modern physics for the derivation) that the change in wavelength of the scattered photon is given by:

𝜆′ − 𝜆 = ℎ

𝑚𝑒𝑐 (1 − cos 𝜃). (7)

This result can be expressed directly in terms of the energies 𝐸′ and 𝐸 of the scattered and unscattered gamma rays, respectively, by using 𝜆 = ℎ𝑐/𝐸:

1

𝐸′ − 1

𝐸=

1

𝑚𝑒𝑐2 (1 − cos 𝜃). (8)

This change of gamma-ray wavelength (or energy) to a different wavelength upon scattering from an electron is called the Compton Effect. The object of this experiment is to measure the energy change of the scattered gamma-ray, and to compare the results with Eq. (7) or (8). The apparatus used in this experiment consists of one strong 137Cs source used in parts 1, 2, and 3. Both its scattered and unscattered energies will be determined. Three weak calibrating sources (137Cs, 22Na, and 60Co) will be used solely for calibrating a pulse-height analyzer. In addition to these sources, there is a sodium iodide detector, a scaler-time-counter, associated power supplies, and a computer that contains a pulse-height analyzer. The geometry of the scatterers is based on a theorem from plane geometry that states that, if a triangle is inscribed inside a circle with a fixed chord of the circle as one

Page 3: Experiment 11 - Physics

Experiment 11

3

side, then the angle opposite the fixed chord is a constant (see Fig. 3). Thus, if a source is placed at one end of the chord, and a photon emitted from this point is scattered

Figure 2: Geometry of the Source, Detector, and the 90° Scattering Plates

Figure 3: If a triangle is inscribed in a circle with a fixed chord of the circle as one side, then

the opposite angle of the triangle is a constant. by electrons placed on the circumference of the circle, and if the photon is subsequently detected by a counter placed at the other end of the chord, then the scattering angle is fixed by the geometry. If a circular plate is shaped to fit the circumference of the circle and the photons are allowed to scatter off the entire plate, then any photon which is detected after one scattering must have been scattered through the same angle.

Page 4: Experiment 11 - Physics

Experiment 11

4

The particular setup used in this experiment has been constructed in such a way that the two different scattering angles may be used: 69° and 90°. These angles were arbitrarily chosen and there is no reason in principle why one could not construct scatterers for any chosen angle. The scatterers used are thin aluminum plates because aluminum has relatively high density of free electrons. Referring to Eq. (7) or (8), the constants ℎ, 𝑚𝑒 , and 𝑐 are known fundamental constants. Therefore, in order to verify the correctness of these predictions, we must measure the energies of both the unscattered (𝐸) and the scattered (𝐸′) photons for one or more particular angles. Since the mass absorption coefficient of lead is a known function of energy, the energy measurement in this experiment is performed by measuring the mass absorption coefficient of the gamma rays in lead. For a given absorbing material, the absorption probability is proportional to the number of atoms of absorber per unit of cross sectional area. Therefore, it is convenient to measure absorber thickness in cm. A series of cylindrical lead absorbers is provided. These are constructed so that they may be placed in successively thicker layers around the detector. A beam of gamma rays of initial intensity 𝐼0 is attenuated upon passing through a thickness 𝑡 (in g/cm2) of absorber according to the equation: 𝐼 = 𝐼0𝑒𝑥𝑝(−𝜇𝑡) (9) where 𝜇 is the mass absorption coefficient in cm2/g. The intensities 𝐼 and 𝐼0 are in terms of number of gamma rays per square centimeter per second. For a particular geometrical configuration of source and scatterer, the mass absorption coefficient is determined by measuring the counting rate (intensity) as a function of the thickness of the absorbers. This functional relationship may be then plotted by using a weighted linear-regression computer program such as Wlinfit.nb or its equivalent. Wlinfit.nb gives both the slope of the resulting straight line and its uncertainty. The energy may then be determined from the curve provided (Fig. 6). The sodium iodide (NaI) scintillation detector used in this experiment is an important and widely used device for detecting gamma rays. As shown in Fig. 4, it consists of a crystal of NaI to which a small amount of thallium has been added in order to make it an effective scintillator. The cylindrical NaI (2 ½ x 1 ¾ inch diameter) is hermetically encased in an aluminum cup with a glass window on one end. The glass window of the detector is mounted on the face of a 10-stage photo-multiplier tube. Gamma rays entering the NaI detector will interact by either the photoelectric or Compton process. Gamma rays with energy greater than 1.02 MeV can also produce an electron-position pair. The photoelectric process results in the gamma ray imparting essentially all of its energy to one of the bound electrons in the crystal. The Compton process is assumed to occur on a free electron, and as we have just seen, results in only a fraction of the gamma-

Page 5: Experiment 11 - Physics

Experiment 11

5

ray energy being carried off by an electron. Electrons from either process will quickly lose their energy in the crystal by causing ionizing events with the atoms of the crystal with the net effect that photons in the visible region will be produced in the de-exitation of the atoms. The number of visible photons produced is directly proportional to the energy deposited by a gamma ray in the crystal. The photons are reflected by a coating of MgO surrounding the crystal except for the end covered by the glass window and many of them enter the photomultiplier.

Figure 4:Gamma-ray detection and counting equipment, configured for parts one and two.

Event (a) is a Compton scattering event and event (b) is a photoelectric process. The photons entering the photomultiplier will strike a photocathode surface present on the inside face of the phototube. Low energy electrons will be ejected from the photocathode and some of them will be incident on the first dynode of the tube. The dynode provides amplification of the number of electrons by the secondary emission process (more electrons are emitted than are incident on a given dynode). The number of electrons is increased by nine successive dynode stages and finally the electrons are collected on the anode where a reasonably large voltage pulse will result. The amplitude of the voltage pulse will again be directly proportional to the energy deposited in the NaI.

Page 6: Experiment 11 - Physics

Experiment 11

6

After going through a preamplifier at the tube base, the voltage pulse is carried via a cable to the scaler on a table outside of the room where the Compton scattering experiment is located. Procedure: Part 1

1. Energy of the Unscattered Gamma Ray The radioactive source used in this part of the experiment is 137Cs. It has a half life of about 30 years and emits a gamma ray with an energy of 0.662 MeV. Before attempting to measure the energy of the Compton scattered gamma rays, an absorption measurement will be made of the direct or unscattered gamma ray in order to become familiar with the techniques.

a. The AC and DC power should be turned on for the two power supplies in the

room with the scattering experiment. The high voltage for the photomultiplier should be +850 volts. No adjustment of either power supply should be necessary. A small lead source holder and collimator is available and this should be placed 6 to 8 inches from the NaI detector with the collimation hole aimed directly at the detector. No cylindrical lead absorbers should be present around the detector for this part of the experiment.

b. Go outside the room and turn on the AC power to the scaler (the switch is located on the upper left hand side on the back) and the oscilloscope. Oscilloscope control positions should be set as specified in the Addendum to the Compton Effect posted with the apparatus. Appropriate settings for the oscilloscope voltage display are obtained by rotating the gray CH 1 VOLTS/DIV knob to “0.2-0.5 Volts/Div”. This value should lie back opposite the notation “1X PROBE” on the front panel. For best time display, the black brackets on the gray A and B TIME/DIV knob should straddle (select) the “2-5 𝜇s” value. It is instructive to observe the pulses coming from the NaI detector on channel one of the oscilloscope, prior to taking data. With correct settings, the pulses should appear on the oscilloscope as depicted in Fig. (5). The abundant, heavy trace, large pulses correspond to all of the energy of the gamma ray being left in the NaI crystal. Approximate the amplitude (in volts) of these large pulses. The spread of smaller pulses beneath the full energy pulse corresponds to Compton processes where the scattered gamma ray escapes the crystal.

c. First a background measurement should then be made using the scaler with the 137Cs source in the lead storage pig on the floor.

Page 7: Experiment 11 - Physics

Experiment 11

7

d. For part one of the experiment we will use the following configuration to measure the energy of the unscattered gamma ray:

Aim the slotted collimator at the detector. You will place the square lead sheets into the rack on the collimator.

e. Remove the 137Cs source from the lead brick on the floor where it is stored and insert it into the lead collimator. Care should be taken to minimize your exposure to the 137Cs source. Always handle it by the opposite end of the rod to which it is attached. Do not stay in the room while you are taking data. None of the small rectangular lead absorbers are needed at this point.

f. With no absorbers in place, use the scaler to determine the counting rate. Enough time should be allotted to accumulate at least 10,000 counts so that the statistical (random) error is no larger than about 1%. Because the counting rates are so high, 10,000 counts can be obtained in about one second. Measurements should then be carried out with 1, 2, 3… absorbers. Use all 10 absorbers to acquire a reasonable number of counts.

Page 8: Experiment 11 - Physics

Experiment 11

8

Figure 5: Oscilloscope display of the photomultiplier pulses caused by incarceration of

gamma rays with NaI crystal

g. Correct the total counting rate by subtracting the average background count from each of the eleven total counts previously obtained. Then plot an absorption curve consisting of the natural log of the corrected counts per second plotted against absorber thickness 𝒕 in g/cm2 (see page 4 for the definition of 𝒕). 𝑡 has a value of 1.79 g/cm2 for each flat absorber. The plot can be made on the computer by use of the Mathematica “Wlinfit” weighted linear least squares fitting program. Wlinfit is available on the desktop of the lab PC’s in the “Scratch” file.

h. Extract the half-thickness 𝒕𝟏/𝟐 using your Wlinfit regression program

summary that gives the slope and intercept values and their uncertainties. The half-thickness is the absorber thickness required to reduce the counting rate by ½ its initial value. Note that 𝑡1/2 = ln(2)/𝑚, where 𝑚 is the slope from Wlinfit. Since 𝜇 =

ln(2)/𝑡1/2, then 𝜇 = 𝑚.

i. With the use of Fig. 6 where the mass absorption coefficient for the lead is

plotted as a function of gamma ray energy, determine the energy of the 137Cs gamma ray from your value of 𝝁. Compare your result with the known value of 0.662 MeV. One can graphically determine the uncertainty in

Page 9: Experiment 11 - Physics

Experiment 11

9

the value of 𝜇 (and therefore energy) using the Wlinfit Mathematica document.

Page 10: Experiment 11 - Physics

Experiment 11

10

Procedure: Part 2

1. Energy of the Scattered Gamma Rays For this part of the experiment, the 137Cs source should be inserted in the source holder on the left-hand side of the Compton scattering table. The cylindrical lead absorbers that are to be used fit directly around the NaI scintillation detector. Place one or two large lead bricks in the brackets on the table between the collimator (without slots) and the detector.

a. Take background readings for the 69° and 90° scattering measurements

by having the source in place but the scatterers removed. This configuration will allow all radiation that reaches the detector but which does not comes from the scatterers to be subtracted as background. The best data is obtained if both background and total counts are taken for each of the five configurations of absorbers shown below. The suggested counting times are:

Number of Cylindrical Absorbers Counting Time

0 1

20 sec 20 sec

3 50 sec 5 60 sec 7 5 min

b. Repeat the measurements discussed in Part 1a. first with the 69° and

then the 90° scattering. By using “Wlinfit” or its equivalent, determine the absorption coefficient, 𝝁, for each angle plotted, followed by their energies from Fig. 6 in the fashion described previously. Graphically estimate the uncertainty in your two energy measurements from your absorption curve. For both scattering angles, first 69° and then 90°, and with no cylindrical absorbers in place, observe the pulses in the Compton band on the oscilloscope and try to estimate their height in volts. These consist of the most intense pulses of greatest amplitude. How do these maximum pulse heights compare to those observed with the unscattered gamma ray? Estimate the energies in MeV of the scattered radiation for both scattering angles, based on the maximum pulse height read for the 137Cs gamma ray that corresponds to 0.662 MeV.

c. Comparison of experimental results with theory

Using Eq. (8), calculate the energy of the scattered gamma ray 𝑬′, with 𝑬 = 0.662 MeV. Do this first with 𝜃 = 69°, and the second with 𝜃 = 90°. Compare the expected values of 𝐸′ with your measured values and discuss any discrepancies. From a linear plot 𝟏/𝑬′ versus (𝟏 − 𝐜𝐨𝐬 𝜽) for the three

Page 11: Experiment 11 - Physics

Experiment 11

11

angles 0°, 69°, and 90°, find the value of 𝒎𝒆𝒄𝟐 and compare it with the established value of 0.511 MeV.

Procedure: Part 3

1. Determination of Energy of the Scattered Gamma Rays Using a Pulse Height Analyzer A pulse height analyzer is a special-purpose computer that sorts and counts pulses. The amplitude of each incoming pulse is measured and is then converted to a digital value that is stored in a memory register location that is a part of a larger (16385 channel) memory. Since the voltage amplitude of each pulse is proportional to the energy of the photon that produced it, the PHA, in effect, assigns energy to channels in memory, in the order of ascending energy. This can be described algebraically by the equation E = mC + Eo, where E is energy and C is a corresponding channel number. It is in the general form y = mx + B, thus E0 is the intercept value of energy and m the slope of this linear relation between energy and channel location in the computer.

In operation, pulses go from the NaI detector into a photo-multiplier tube, through an amplifier, and then into the PHA (see Fig. 7). The PHA then stores them over time and constructs a frequency distribution of the incoming pulses plotted against channel number. It then displays the distribution as a histogram on the computer screen with total counts per channel on the y-axis and channel numbers on the x-axis. If the pulses resulting from a 137Cs gamma ray are sorted according to their height, a spectrum will be obtained similar to the one shown in Fig. 8. In Fig. 8, the peak on the right is the full energy peak and corresponds to the 0.662 MeV gamma ray having a photoelectric event in the crystal or a Compton event where the scattered photon is also stopped in the crystal. This full energy or photoelectric peak is a useful calibration marker for measuring the energy of unknown gamma rays. The rather flat bump to the left of the full energy peak corresponds mostly to Compton events in the crystal where various amounts of energy escape from the crystal. The small peak at the far left is the photo-peak of a 32 keV x-ray that also comes from the 137Cs radioactive decay.

Page 12: Experiment 11 - Physics

Experiment 11

12

Figure 7: Gamma-ray detection and counting equipment, configured for Part 3. The event

marked (a) is a Compton scattering event and the event marked (b) is a photoelectric process.

Page 13: Experiment 11 - Physics

Experiment 11

13

Figure 8: A Pulse Height Spectrum for a 137Cs Source

2. Energy Calibration of the Scintillation Detector

a. The AC and DC power should be turned on for the two power supplies in the room with the scattering equipment. The high voltage for the photomultiplier should be set to +850 volts. On the table outside the room, turn on the power for the instruments contained in the PORTABLE BIN/POWER SUPPLY. This switch is located on the BIN’s upper left-hand corner on the back. Your lab instructor or the lab coordinator will provide you with three small plastic calibration sources of 137Cs, 22Na, and 60Co. They will also help you to set up the computer so it is ready for use in obtaining pulse height spectra. Place the small plastic 137Cs source in the NaI detector.

b. The oscilloscope should now show the pulses from the 137Cs source as is shown in Fig. 5. The abundant large pulses correspond to the full energy peak (0.662 MeV) and the enhanced pulses near the baseline, the x-ray line at 32 keV.

c. To display energy distributions on the computer screen, proceed as follows:

i. Turn power on to both the Dell PC and its monitor.

Page 14: Experiment 11 - Physics

Experiment 11

14

ii. When the Windows Logon window appears, type “phys2150” in the password box and then press ENTER. The computer will boot to Windows XP desktop.

iii. Now, click the following buttons in order: START ALL PROGRAMS

GENIE 2000 GAMMA ACQUISTION AND ANALYSIS. The Acquisition window will open. Click FILE OPEN DATA SOURCE. Now click the DETECTOR button. A red and yellow radiation symbol labeled “Detector MP2_MCA1” will appear in this window. Click CLEAR START, and you will see that energy peaks begin developing across the bottom of black display screen.

iv. The scroll bar on the right side of the display screen allows one to set

the heights of peaks thus displayed. Move it up or down for best display of heights of energy histograms.

d. The notations “Channel” (in memory) and “Counts” are displayed in blue at

the top of the energy display window. To determine total counts in a given channel for a histogram, click anywhere in it with the mouse pointer. This will select a single memory location and a white marker will appear in the site selected. The channel number and total counts in this particular location will be given at the top. Pressing the left and right keys on the keyboard will cause the marker to scan across the screen to any of the 16384 memory locations.

e. In general, to produce energy peak histograms, click STOP CLEAR START, and then wait as long as is reasonably possible for peaks to develop. Move the marker to the center of peaks of interest and record channel number and observed counts. Counting precision depends on having counts as high as possible. Place the 137Cs source on top of the NaI detector.

f. Follow the steps given previously to determine the appropriate channel

numbers for the 137Cs source. Then, remove the 137Cs source from the NaI crystal detector and repeat this procedure for the 22Na source, placed on top of the detector. Two prominent gamma energies are produced by the 22Na source: a 0.511 MeV line, caused by positron-electron annihilation. That produces two quanta each of 0.511 MeV. The second line is a 1.28 MeV gamma. Remove the 22Na source and replace it with the 60Co source. This produces two high-energy peaks of interest in this experiment. They are 1.17 MeV and 1.33 MeV. Because each separate peak is hard to distinguish, their channel numbers can be determined and the channel numbers averaged, and the average 60Co energy and corresponding average channel number can be used in calibration.

g. Make a calibration graph of the energy (in MeV) of all the data you have

taken. That is, plot energy in MeV against corresponding channel number for

Page 15: Experiment 11 - Physics

Experiment 11

15

the three sources. Also use the Linfit or an equivalent liner least squares fitting program and your data to determine the slope and intercept of the line 𝑬 = 𝒃𝑪 + 𝑬𝟎.

3. Determination of the Energy of the Scattered Gamma Ray

a. Remove the strong 137Cs source from the lead shielding blocks on the floor

and insert it into the hole in the lead brick mounted on the left-hand side of the scattering apparatus table. You should still have one or two lead bricks in the brackets in the middle of the table between the source and detector. Put the 90° aluminum scatterers in their appropriate slots. Run the Acquisition program long enough to establish a well-defined peak and record its channel number.

b. Repeat the process with the 69° scatterers in place.

c. Using your calibration graph (from 1g. above) and the computer best-fit line, determine the energy of the Compton scattered gamma rays and estimate the uncertainty in their energies.

d. Compare the expected values previously obtained using Eq. (8) with

your experimental values and discuss any discrepancies.