Compton Scattering (2) Notes | EduRev

: Compton Scattering (2) Notes | EduRev

 Page 1


30 Sep 09 Compton.1 
COMPTON SCATTERING 
 
 Compton scattering is another phenomenon of electromagnetic radiation which cannot be 
explained in terms of the wave theory.  In this experiment an attempt will be made to verify the 
Compton scattering formula, derived from the quantum theory of electromagnetic radiation, and 
as a consequence, the mass of the electron will be determined. 
 
Theory: 
 
 In 1920, A.H. Compton investigated the scattering of monochromatic x-rays 
(electromagnetic radiation) from various materials.  He observed that after the scattering the 
frequency (energy) of the x-rays had changed, and had always decreased.  From the point of 
view of classical (wave) electromagnetic theory, this frequency shift cannot be explained since 
the frequency is a property of the incoming electromagnetic wave and cannot be altered by the 
change of direction implied by the scattering.  If, on the other hand, the incoming radiation is 
thought of as a beam of photons (electromagnetic quanta) then the situation becomes that of 
photons of energy E = h ? scattering from free electrons in the target material.  Energy-
momentum conservation, applied to this situation, predicts that the scattered photons will have 
energy E ? = h ? ? < E, in complete agreement with Compton’s experiments. 
 
 The frequency shift will depend on the angle of scattering, and can be calculated from 
kinematics.  Consider an incoming photon of energy h ? and momentum h ?/c scattering from any 
electron of mass m.  p is the momentum of the electron after scattering and h ? ?, h? ?/c are the 
energy and momentum of the scattered photon.  For momentum conservation, the three vectors 
h ?/c, h? ?/c, and p must lie in the same plane (see Figure 1). 
 
 
 
Figure 1 
 
From energy conservation, 
 
    hmc h mc pc ?? ?? ???
22422
 (1) 
 
where m is the rest mass of the electron. 
 
Page 2


30 Sep 09 Compton.1 
COMPTON SCATTERING 
 
 Compton scattering is another phenomenon of electromagnetic radiation which cannot be 
explained in terms of the wave theory.  In this experiment an attempt will be made to verify the 
Compton scattering formula, derived from the quantum theory of electromagnetic radiation, and 
as a consequence, the mass of the electron will be determined. 
 
Theory: 
 
 In 1920, A.H. Compton investigated the scattering of monochromatic x-rays 
(electromagnetic radiation) from various materials.  He observed that after the scattering the 
frequency (energy) of the x-rays had changed, and had always decreased.  From the point of 
view of classical (wave) electromagnetic theory, this frequency shift cannot be explained since 
the frequency is a property of the incoming electromagnetic wave and cannot be altered by the 
change of direction implied by the scattering.  If, on the other hand, the incoming radiation is 
thought of as a beam of photons (electromagnetic quanta) then the situation becomes that of 
photons of energy E = h ? scattering from free electrons in the target material.  Energy-
momentum conservation, applied to this situation, predicts that the scattered photons will have 
energy E ? = h ? ? < E, in complete agreement with Compton’s experiments. 
 
 The frequency shift will depend on the angle of scattering, and can be calculated from 
kinematics.  Consider an incoming photon of energy h ? and momentum h ?/c scattering from any 
electron of mass m.  p is the momentum of the electron after scattering and h ? ?, h? ?/c are the 
energy and momentum of the scattered photon.  For momentum conservation, the three vectors 
h ?/c, h? ?/c, and p must lie in the same plane (see Figure 1). 
 
 
 
Figure 1 
 
From energy conservation, 
 
    hmc h mc pc ?? ?? ???
22422
 (1) 
 
where m is the rest mass of the electron. 
 
30 Sep 09 Compton.2 
From momentum conservation, 
 
    
h
c
h
c
p
? ?
? ?
?
? cos cos? (2) 
and 
    0=
h
c
p
?
?
?
? sin sin? (3) 
 
where ? is the photon scattering angle and ? is the electron recoil angle.  Solving (2) and (3) for 
pc cos ? and pc sin ?  respectively, and squaring, 
 
    pc h h h
22 2 2 2 2 2 2 2
2 cos cos cos ? ? ? ? ? ? ? ? ? ? ? ? (4) 
 
    pc h
22 2 2 2 2
sin sin ? ? ? ? ? (5) 
 
Adding these two equations and using sin
2
 + cos
2
 = 1, 
 
    pc h h h
22 2 2 2 2 2
2 ?? ? ? ? ? ? ? ? ?cos (6) 
 
Squaring equation (1) and solving for p
2
c
2
 yields 
 
    pc h h hmc h
22 2 2 2 2 2 2
22 ?? ? ? ? ? ? ? ? ? ? ? ? ? () (7) 
 
Subtracting equations (6) and (7), 
 
    
? ?
? ?
?
? ?
?
??
h
mc
2
1 (cos ) (8) 
 
Converting frequency to wavelength ( ? = c/ ?) gives 
 
    ??? ? ? ? ??? ?
h
mc
(cos 1 ) (9) 
 
for the shift in wavelength of the scattered beam, while converting frequency to energy ( ? = E/h) 
yields 
 
    ? ?
??
E
E
E
mc
11
2
(cos ?)
 (10) 
 
for the energy of the scattered photons.  Re-arranging equation (10) yields 
 
    
11 1
1
2
?
?? ?
E E mc
(cos ?) (11) 
where E is the incoming photon energy and E ? is the scattered photon energy.  Thus the Compton 
formula, equation (11), predicts that the inverse of the energy of the scattered photon varies 
linearly as (1 – cos ?) where ? is the photon scattering angle. 
Page 3


30 Sep 09 Compton.1 
COMPTON SCATTERING 
 
 Compton scattering is another phenomenon of electromagnetic radiation which cannot be 
explained in terms of the wave theory.  In this experiment an attempt will be made to verify the 
Compton scattering formula, derived from the quantum theory of electromagnetic radiation, and 
as a consequence, the mass of the electron will be determined. 
 
Theory: 
 
 In 1920, A.H. Compton investigated the scattering of monochromatic x-rays 
(electromagnetic radiation) from various materials.  He observed that after the scattering the 
frequency (energy) of the x-rays had changed, and had always decreased.  From the point of 
view of classical (wave) electromagnetic theory, this frequency shift cannot be explained since 
the frequency is a property of the incoming electromagnetic wave and cannot be altered by the 
change of direction implied by the scattering.  If, on the other hand, the incoming radiation is 
thought of as a beam of photons (electromagnetic quanta) then the situation becomes that of 
photons of energy E = h ? scattering from free electrons in the target material.  Energy-
momentum conservation, applied to this situation, predicts that the scattered photons will have 
energy E ? = h ? ? < E, in complete agreement with Compton’s experiments. 
 
 The frequency shift will depend on the angle of scattering, and can be calculated from 
kinematics.  Consider an incoming photon of energy h ? and momentum h ?/c scattering from any 
electron of mass m.  p is the momentum of the electron after scattering and h ? ?, h? ?/c are the 
energy and momentum of the scattered photon.  For momentum conservation, the three vectors 
h ?/c, h? ?/c, and p must lie in the same plane (see Figure 1). 
 
 
 
Figure 1 
 
From energy conservation, 
 
    hmc h mc pc ?? ?? ???
22422
 (1) 
 
where m is the rest mass of the electron. 
 
30 Sep 09 Compton.2 
From momentum conservation, 
 
    
h
c
h
c
p
? ?
? ?
?
? cos cos? (2) 
and 
    0=
h
c
p
?
?
?
? sin sin? (3) 
 
where ? is the photon scattering angle and ? is the electron recoil angle.  Solving (2) and (3) for 
pc cos ? and pc sin ?  respectively, and squaring, 
 
    pc h h h
22 2 2 2 2 2 2 2
2 cos cos cos ? ? ? ? ? ? ? ? ? ? ? ? (4) 
 
    pc h
22 2 2 2 2
sin sin ? ? ? ? ? (5) 
 
Adding these two equations and using sin
2
 + cos
2
 = 1, 
 
    pc h h h
22 2 2 2 2 2
2 ?? ? ? ? ? ? ? ? ?cos (6) 
 
Squaring equation (1) and solving for p
2
c
2
 yields 
 
    pc h h hmc h
22 2 2 2 2 2 2
22 ?? ? ? ? ? ? ? ? ? ? ? ? ? () (7) 
 
Subtracting equations (6) and (7), 
 
    
? ?
? ?
?
? ?
?
??
h
mc
2
1 (cos ) (8) 
 
Converting frequency to wavelength ( ? = c/ ?) gives 
 
    ??? ? ? ? ??? ?
h
mc
(cos 1 ) (9) 
 
for the shift in wavelength of the scattered beam, while converting frequency to energy ( ? = E/h) 
yields 
 
    ? ?
??
E
E
E
mc
11
2
(cos ?)
 (10) 
 
for the energy of the scattered photons.  Re-arranging equation (10) yields 
 
    
11 1
1
2
?
?? ?
E E mc
(cos ?) (11) 
where E is the incoming photon energy and E ? is the scattered photon energy.  Thus the Compton 
formula, equation (11), predicts that the inverse of the energy of the scattered photon varies 
linearly as (1 – cos ?) where ? is the photon scattering angle. 
30 Sep 09 Compton.3 
Apparatus: 
 The source of photons (electromagnetic radiation) for this experiment is 50 mCi of 
137
Cs.  
137
Cs emits 0.662 MeV gamma rays.  Although Compton’s experiments were done with x-rays, a 
gamma ray source should work just as well, since both x-rays and gamma rays are forms of 
electromagnetic radiation.  The source is placed inside a lead collimator.  DO NOT REMOVE 
THE SOURCE FROM THE LEAD COLLIMATOR.  EXPOSURE TO THE RADIATION 
WILL BE HAZARDOUS.  SPEND AS LITTLE TIME AS NECESSARY STANDING IN 
FRONT OF THE SOURCE OPENING. 
 The scattering target is a 1 cm diameter aluminum rod.  The rod may be removed by 
unscrewing it from the baseplate. 
 The detection and analysis system consists of a NaI(Tl) scintillation crystal and 
photomultiplier tube, a scintillation amplifier/high voltage power supply, and a personal 
computer with a multi-channel analyser card.  Gamma rays passing into the NaI(Tl) crystal cause 
flashes of light (scintillations) inside the crystal.  These flashes of light release electrons from the 
photocathode of the photomultiplier tube (by the photoelectric effect).  The high voltage applied 
to the photomultiplier tube causes the electrons to be channelled through the various stages of 
the tube, with amplification of the number of electrons occurring at each stage.  The result is a 
pulse at the output of the photomultiplier tube, the voltage of the pulse being proportional to the 
energy of the gamma ray incident on the crystal (provided the gamma ray stops in the crystal).  
After linear amplification the voltage pulse is digitized by the 'analog to digital converter' (ADC) 
card in the computer.  The monitor displays the number of pulses versus channel number.  The 
channel number is directly proportional to the photomultiplier tube pulse voltage and hence to 
the deposited gamma ray energy.  The monitor thus shows the energy distribution of the gamma 
rays being detected. 
 The scintillation crystal/photomultiplier tube assembly is mounted inside a heavy lead shield 
to reduce background radiation.  The assembly, with shielding, is mounted on a trolley which 
can be rotated about a vertical axis centred on the target.  The scattering angle can be measured 
on a protractor attached to the baseplate.  Figure 2 is a diagram of the apparatus. 
Page 4


30 Sep 09 Compton.1 
COMPTON SCATTERING 
 
 Compton scattering is another phenomenon of electromagnetic radiation which cannot be 
explained in terms of the wave theory.  In this experiment an attempt will be made to verify the 
Compton scattering formula, derived from the quantum theory of electromagnetic radiation, and 
as a consequence, the mass of the electron will be determined. 
 
Theory: 
 
 In 1920, A.H. Compton investigated the scattering of monochromatic x-rays 
(electromagnetic radiation) from various materials.  He observed that after the scattering the 
frequency (energy) of the x-rays had changed, and had always decreased.  From the point of 
view of classical (wave) electromagnetic theory, this frequency shift cannot be explained since 
the frequency is a property of the incoming electromagnetic wave and cannot be altered by the 
change of direction implied by the scattering.  If, on the other hand, the incoming radiation is 
thought of as a beam of photons (electromagnetic quanta) then the situation becomes that of 
photons of energy E = h ? scattering from free electrons in the target material.  Energy-
momentum conservation, applied to this situation, predicts that the scattered photons will have 
energy E ? = h ? ? < E, in complete agreement with Compton’s experiments. 
 
 The frequency shift will depend on the angle of scattering, and can be calculated from 
kinematics.  Consider an incoming photon of energy h ? and momentum h ?/c scattering from any 
electron of mass m.  p is the momentum of the electron after scattering and h ? ?, h? ?/c are the 
energy and momentum of the scattered photon.  For momentum conservation, the three vectors 
h ?/c, h? ?/c, and p must lie in the same plane (see Figure 1). 
 
 
 
Figure 1 
 
From energy conservation, 
 
    hmc h mc pc ?? ?? ???
22422
 (1) 
 
where m is the rest mass of the electron. 
 
30 Sep 09 Compton.2 
From momentum conservation, 
 
    
h
c
h
c
p
? ?
? ?
?
? cos cos? (2) 
and 
    0=
h
c
p
?
?
?
? sin sin? (3) 
 
where ? is the photon scattering angle and ? is the electron recoil angle.  Solving (2) and (3) for 
pc cos ? and pc sin ?  respectively, and squaring, 
 
    pc h h h
22 2 2 2 2 2 2 2
2 cos cos cos ? ? ? ? ? ? ? ? ? ? ? ? (4) 
 
    pc h
22 2 2 2 2
sin sin ? ? ? ? ? (5) 
 
Adding these two equations and using sin
2
 + cos
2
 = 1, 
 
    pc h h h
22 2 2 2 2 2
2 ?? ? ? ? ? ? ? ? ?cos (6) 
 
Squaring equation (1) and solving for p
2
c
2
 yields 
 
    pc h h hmc h
22 2 2 2 2 2 2
22 ?? ? ? ? ? ? ? ? ? ? ? ? ? () (7) 
 
Subtracting equations (6) and (7), 
 
    
? ?
? ?
?
? ?
?
??
h
mc
2
1 (cos ) (8) 
 
Converting frequency to wavelength ( ? = c/ ?) gives 
 
    ??? ? ? ? ??? ?
h
mc
(cos 1 ) (9) 
 
for the shift in wavelength of the scattered beam, while converting frequency to energy ( ? = E/h) 
yields 
 
    ? ?
??
E
E
E
mc
11
2
(cos ?)
 (10) 
 
for the energy of the scattered photons.  Re-arranging equation (10) yields 
 
    
11 1
1
2
?
?? ?
E E mc
(cos ?) (11) 
where E is the incoming photon energy and E ? is the scattered photon energy.  Thus the Compton 
formula, equation (11), predicts that the inverse of the energy of the scattered photon varies 
linearly as (1 – cos ?) where ? is the photon scattering angle. 
30 Sep 09 Compton.3 
Apparatus: 
 The source of photons (electromagnetic radiation) for this experiment is 50 mCi of 
137
Cs.  
137
Cs emits 0.662 MeV gamma rays.  Although Compton’s experiments were done with x-rays, a 
gamma ray source should work just as well, since both x-rays and gamma rays are forms of 
electromagnetic radiation.  The source is placed inside a lead collimator.  DO NOT REMOVE 
THE SOURCE FROM THE LEAD COLLIMATOR.  EXPOSURE TO THE RADIATION 
WILL BE HAZARDOUS.  SPEND AS LITTLE TIME AS NECESSARY STANDING IN 
FRONT OF THE SOURCE OPENING. 
 The scattering target is a 1 cm diameter aluminum rod.  The rod may be removed by 
unscrewing it from the baseplate. 
 The detection and analysis system consists of a NaI(Tl) scintillation crystal and 
photomultiplier tube, a scintillation amplifier/high voltage power supply, and a personal 
computer with a multi-channel analyser card.  Gamma rays passing into the NaI(Tl) crystal cause 
flashes of light (scintillations) inside the crystal.  These flashes of light release electrons from the 
photocathode of the photomultiplier tube (by the photoelectric effect).  The high voltage applied 
to the photomultiplier tube causes the electrons to be channelled through the various stages of 
the tube, with amplification of the number of electrons occurring at each stage.  The result is a 
pulse at the output of the photomultiplier tube, the voltage of the pulse being proportional to the 
energy of the gamma ray incident on the crystal (provided the gamma ray stops in the crystal).  
After linear amplification the voltage pulse is digitized by the 'analog to digital converter' (ADC) 
card in the computer.  The monitor displays the number of pulses versus channel number.  The 
channel number is directly proportional to the photomultiplier tube pulse voltage and hence to 
the deposited gamma ray energy.  The monitor thus shows the energy distribution of the gamma 
rays being detected. 
 The scintillation crystal/photomultiplier tube assembly is mounted inside a heavy lead shield 
to reduce background radiation.  The assembly, with shielding, is mounted on a trolley which 
can be rotated about a vertical axis centred on the target.  The scattering angle can be measured 
on a protractor attached to the baseplate.  Figure 2 is a diagram of the apparatus. 
30 Sep 09 Compton.4 
Figure 2 
 
 
Procedure and Experiment: 
 
NOTE: Refer to Figure 2 for the definition of the scattering angle, ?, and note that the labelling 
of the protractor corresponds to 90° – ?.  i.e. ? = 90° – protractor reading. 
 
 Before the actual Compton scattering experiment can be performed, two calibrations are 
required.  First, the conversion equation between channel number and energy is determined; 
second, the protractor reading corresponding to ? = 0° is found. 
 
1. Log-in to the computer (there is no password). 
2. Turn on the UCS30 device. 
3. Double-click on the UCS30 icon on the desktop. 
4. Click on the Mode menu and select ‘PHA (Amp In)’, the 2
nd
 item on the list. 
5. Click on the Settings menu and select Amp/HV/ADC.  The high voltage should be set to 450. 
Click the button to turn it On.  Check that the coarse and fine amplifier gains are 32 and 1 
respectively, the conversion gain is 2048, and that the Amp In Polarity is positive. (If any of 
these settings are different than stated above, consult the lab instructor.) 
6. Data acquisition is controlled using the buttons on the toolbar.  For the most part, the 
functions of these buttons are obvious, and a ToolTip will appear if you hover over a button. 
Data acquisition is begun by clicking the ‘GO’ button. 
Data acquisition is stopped by clicking the ‘STOP’ button. 
Data is deleted using the eraser button (3
rd
 from left). 
Data can be acquired for a pre-set time by clicking the clock button.  Ensure that ‘Live Time’ 
is selected. 
To determine the total number of counts in some region, or to determine the channel 
number/energy of a peak in the spectrum, a region of interest (ROI) must be defined.  This is 
done using the ROI button.  A ROI is defined by placing the cursor at the left edge of the 
desired region and then clicking with the left mouse button and dragging the cursor to the 
Page 5


30 Sep 09 Compton.1 
COMPTON SCATTERING 
 
 Compton scattering is another phenomenon of electromagnetic radiation which cannot be 
explained in terms of the wave theory.  In this experiment an attempt will be made to verify the 
Compton scattering formula, derived from the quantum theory of electromagnetic radiation, and 
as a consequence, the mass of the electron will be determined. 
 
Theory: 
 
 In 1920, A.H. Compton investigated the scattering of monochromatic x-rays 
(electromagnetic radiation) from various materials.  He observed that after the scattering the 
frequency (energy) of the x-rays had changed, and had always decreased.  From the point of 
view of classical (wave) electromagnetic theory, this frequency shift cannot be explained since 
the frequency is a property of the incoming electromagnetic wave and cannot be altered by the 
change of direction implied by the scattering.  If, on the other hand, the incoming radiation is 
thought of as a beam of photons (electromagnetic quanta) then the situation becomes that of 
photons of energy E = h ? scattering from free electrons in the target material.  Energy-
momentum conservation, applied to this situation, predicts that the scattered photons will have 
energy E ? = h ? ? < E, in complete agreement with Compton’s experiments. 
 
 The frequency shift will depend on the angle of scattering, and can be calculated from 
kinematics.  Consider an incoming photon of energy h ? and momentum h ?/c scattering from any 
electron of mass m.  p is the momentum of the electron after scattering and h ? ?, h? ?/c are the 
energy and momentum of the scattered photon.  For momentum conservation, the three vectors 
h ?/c, h? ?/c, and p must lie in the same plane (see Figure 1). 
 
 
 
Figure 1 
 
From energy conservation, 
 
    hmc h mc pc ?? ?? ???
22422
 (1) 
 
where m is the rest mass of the electron. 
 
30 Sep 09 Compton.2 
From momentum conservation, 
 
    
h
c
h
c
p
? ?
? ?
?
? cos cos? (2) 
and 
    0=
h
c
p
?
?
?
? sin sin? (3) 
 
where ? is the photon scattering angle and ? is the electron recoil angle.  Solving (2) and (3) for 
pc cos ? and pc sin ?  respectively, and squaring, 
 
    pc h h h
22 2 2 2 2 2 2 2
2 cos cos cos ? ? ? ? ? ? ? ? ? ? ? ? (4) 
 
    pc h
22 2 2 2 2
sin sin ? ? ? ? ? (5) 
 
Adding these two equations and using sin
2
 + cos
2
 = 1, 
 
    pc h h h
22 2 2 2 2 2
2 ?? ? ? ? ? ? ? ? ?cos (6) 
 
Squaring equation (1) and solving for p
2
c
2
 yields 
 
    pc h h hmc h
22 2 2 2 2 2 2
22 ?? ? ? ? ? ? ? ? ? ? ? ? ? () (7) 
 
Subtracting equations (6) and (7), 
 
    
? ?
? ?
?
? ?
?
??
h
mc
2
1 (cos ) (8) 
 
Converting frequency to wavelength ( ? = c/ ?) gives 
 
    ??? ? ? ? ??? ?
h
mc
(cos 1 ) (9) 
 
for the shift in wavelength of the scattered beam, while converting frequency to energy ( ? = E/h) 
yields 
 
    ? ?
??
E
E
E
mc
11
2
(cos ?)
 (10) 
 
for the energy of the scattered photons.  Re-arranging equation (10) yields 
 
    
11 1
1
2
?
?? ?
E E mc
(cos ?) (11) 
where E is the incoming photon energy and E ? is the scattered photon energy.  Thus the Compton 
formula, equation (11), predicts that the inverse of the energy of the scattered photon varies 
linearly as (1 – cos ?) where ? is the photon scattering angle. 
30 Sep 09 Compton.3 
Apparatus: 
 The source of photons (electromagnetic radiation) for this experiment is 50 mCi of 
137
Cs.  
137
Cs emits 0.662 MeV gamma rays.  Although Compton’s experiments were done with x-rays, a 
gamma ray source should work just as well, since both x-rays and gamma rays are forms of 
electromagnetic radiation.  The source is placed inside a lead collimator.  DO NOT REMOVE 
THE SOURCE FROM THE LEAD COLLIMATOR.  EXPOSURE TO THE RADIATION 
WILL BE HAZARDOUS.  SPEND AS LITTLE TIME AS NECESSARY STANDING IN 
FRONT OF THE SOURCE OPENING. 
 The scattering target is a 1 cm diameter aluminum rod.  The rod may be removed by 
unscrewing it from the baseplate. 
 The detection and analysis system consists of a NaI(Tl) scintillation crystal and 
photomultiplier tube, a scintillation amplifier/high voltage power supply, and a personal 
computer with a multi-channel analyser card.  Gamma rays passing into the NaI(Tl) crystal cause 
flashes of light (scintillations) inside the crystal.  These flashes of light release electrons from the 
photocathode of the photomultiplier tube (by the photoelectric effect).  The high voltage applied 
to the photomultiplier tube causes the electrons to be channelled through the various stages of 
the tube, with amplification of the number of electrons occurring at each stage.  The result is a 
pulse at the output of the photomultiplier tube, the voltage of the pulse being proportional to the 
energy of the gamma ray incident on the crystal (provided the gamma ray stops in the crystal).  
After linear amplification the voltage pulse is digitized by the 'analog to digital converter' (ADC) 
card in the computer.  The monitor displays the number of pulses versus channel number.  The 
channel number is directly proportional to the photomultiplier tube pulse voltage and hence to 
the deposited gamma ray energy.  The monitor thus shows the energy distribution of the gamma 
rays being detected. 
 The scintillation crystal/photomultiplier tube assembly is mounted inside a heavy lead shield 
to reduce background radiation.  The assembly, with shielding, is mounted on a trolley which 
can be rotated about a vertical axis centred on the target.  The scattering angle can be measured 
on a protractor attached to the baseplate.  Figure 2 is a diagram of the apparatus. 
30 Sep 09 Compton.4 
Figure 2 
 
 
Procedure and Experiment: 
 
NOTE: Refer to Figure 2 for the definition of the scattering angle, ?, and note that the labelling 
of the protractor corresponds to 90° – ?.  i.e. ? = 90° – protractor reading. 
 
 Before the actual Compton scattering experiment can be performed, two calibrations are 
required.  First, the conversion equation between channel number and energy is determined; 
second, the protractor reading corresponding to ? = 0° is found. 
 
1. Log-in to the computer (there is no password). 
2. Turn on the UCS30 device. 
3. Double-click on the UCS30 icon on the desktop. 
4. Click on the Mode menu and select ‘PHA (Amp In)’, the 2
nd
 item on the list. 
5. Click on the Settings menu and select Amp/HV/ADC.  The high voltage should be set to 450. 
Click the button to turn it On.  Check that the coarse and fine amplifier gains are 32 and 1 
respectively, the conversion gain is 2048, and that the Amp In Polarity is positive. (If any of 
these settings are different than stated above, consult the lab instructor.) 
6. Data acquisition is controlled using the buttons on the toolbar.  For the most part, the 
functions of these buttons are obvious, and a ToolTip will appear if you hover over a button. 
Data acquisition is begun by clicking the ‘GO’ button. 
Data acquisition is stopped by clicking the ‘STOP’ button. 
Data is deleted using the eraser button (3
rd
 from left). 
Data can be acquired for a pre-set time by clicking the clock button.  Ensure that ‘Live Time’ 
is selected. 
To determine the total number of counts in some region, or to determine the channel 
number/energy of a peak in the spectrum, a region of interest (ROI) must be defined.  This is 
done using the ROI button.  A ROI is defined by placing the cursor at the left edge of the 
desired region and then clicking with the left mouse button and dragging the cursor to the 
30 Sep 09 Compton.5 
right edge of the region.  The ROI will change colour to show the region that has been 
selected.  The total number of counts in the region is given by the GROSS number (not the 
NET number).  The CENTROID is the channel number or energy of the peak within the ROI.  
To clear the ROI, go to Settings ? ROIs. 
The vertical axis can be toggled between a linear or logarithmic scale with the ‘Y/lin’ and 
‘Y/log’ buttons.  The vertical axis scale is changed by moving the slider at the far right side 
of the program window. 
6. Allow five minutes for the high voltage power supply to warm up. 
7. The system is now ready for use. 
Energy Calibration 
 A number of plastic disk sources of known gamma energies are supplied.  Determining the 
peak channel numbers for these known energies allows calibration of the equipment. 
Source Energy (KeV) 
22
Na 511 
54
Mn 835 
57
Co 122 
 To avoid contamination of the data due to radiation from the large source, be sure the plug is 
in place, remove the scattering target, and rotate the trolley away from the incident direction.  
The plastic sources can be taped over the opening in the detector shielding.  Do not put the tape 
on the label side of the disk sources! 
 Begin data acquisition of the gamma ray spectrum for the 
22
Na disk source and allow to 
continue until there is a clearly identifiable peak on the monitor display. 
Define an appropriate ROI that brackets the gamma ray peak and record the peak channel 
number (Centroid) and FWHM (half the FWHM is the error in the peak channel number). 
DO NOT erase the 
22
Na spectrum – repeat the process of determining the peak channel number 
(centroid) and FWHM for the 
54
Mn and 
57
Co sources.  You should now have three ROIs defining 
the three calibration peaks. 
 Calibration is done as follows: 
? From the Settings menu select Energy Calibrate and 3 point. 
? If prompted, enter the units as KeV. 
? Put the cursor in the ROI of one of the peaks, enter the corresponding energy (in KeV), 
and click OK. 
? Repeat for the other two peaks. 
Once the calibration procedure has been completed, the horizontal axis of the display will read 
energy in KeV. 
As a check of the linearity of the system’s response, plot by hand the energy calibration curve, 
i.e. plot gamma ray energy vs. peak channel number.  The equation of this line allows manual 
conversion from channel number to energy. 
 
Protractor Calibration 
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