nucp 2371 radiation measurements ii. since electromagnetic radiation does not cause any direct...
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Gamma SpectroscopyNUCP 2371
Radiation Measurements II
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Since electromagnetic radiation does not cause any direct ionization
Detector system needs to do several things◦ Reasonable probability to convert gamma rays to
electrons◦ Must be able to detect these electrons◦ Able to separate signal output from detector
Spectrometer
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Since EM does not have any charge it must interact directly with the electron to cause ionization
The more electrons to interact with the higher the probability to create an ion pair
Usually if have more electrons will have more Protons
More protons usually (but not always) leads to a higher density
Density is main characteristic of how EM radiation will interact with mater
EM radiation interactions
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Rayleigh scattering- scattering of light by very tiny particles not very important because it does not cause ionization
Photoelectric effect- electron scattering Comptom scattering- electron scattering
and an extra gamma ray Pair Production- creation of two charged
particles
EM Interactions
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Low energy process Electron absorbs all energy of incoming photon Electron gets ejected with the energy of the
photon minus the binding energy of the electron
Binding energies are several to tens of eV Kinetic Energy of all electrons should be the
same as the incoming photon If all energy is captured in detector will get a
single peak corresponding to the photons energy
Photoelectric Interactions
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Medium energy Electron absorbs some the incoming
photons energy and creates a new lower energy photon
Energy distribution is based on the scattered angle
All scattered angles will be in detector so electron will have a series of energies
Compton Interactions
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Leads to the Compton continuum, the misc energies of the photon that gets scattered at different angles which lead to different energy deposition in the detector
Compton edge distance from photopeak is dependant of energy of photon but is usually limited to 256 keV
Compton (cont)
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λ1- λ0= h/mec (1- cosθ)
λ1= wave after scattering λ0 = initial wave of EM radiation h/mec= Compton wavelength of electron
2.43 X 10 -12 m h=planks constant m= mass of electron c= speed of light
θ = angle of scatter Energy= h f E= hc/ λ because λf=c h= planks constant= 4.13 E-15 eV s
Compton Scattering
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Determine λ of initial EM radiationCalculate change in λAdd change in λ to original λThen calculate the new energy of the scatter
EM radiation Start with 500 keV photon that is deflected
45degrees What next?
Energy change
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High energy process High energy EM interacts with the strong
electric field of the nucleus Produces an electron and positron Energy of the two particles is energy of the
EM wave minus 1.022MeV (rest mass energy of a β- and β +)
Responsible for single and double escape peaks if energy of these particles escapes the detector
Pair Production
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The size of the signals generated from scintillators and solid state detectors are proportional to the energy deposited in the crystal.
Signals can be organized with respect to energy
Each time a signal of a certain magnitude is counted it is added to that energy category
So more of the same size signals that get produced from the detector will result in a larger peak at that corresponding energy
Result will be a chart (spectrum) of the energies of the gamma rays and their intensities
Gamma Spectroscopy
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Can be used to identify unknown radionuclides
Can be used determine quantity of material present when calibrated
Can be quite complicated if have many radionuclides present
Gamma Spectroscopy
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NaI◦ Higher efficiency◦ Lower cost◦ Larger sizes
HPGE◦ Better resolution◦ Better for detecting weak sources
But if need resolution in complex spectra no better than HPGe
NaI vs HPGE
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NaI and Ge detectors
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Uranium Spectrum
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Photo peak Xrays Compton edge Single escape peak Double escape peak Sum peak 511 peak Ge escape peak FWHM
Spectrum
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Large peak in the spectrum that correlates to the full energy of the EM radiation
Ideally would be straight line if all energy from EM would be collected in detector
Is a wider peak due to some of the energy of the EM is lost from the detector
Photo Peak
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Photopeak
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Low energy peaks that corresponds to either characteristic X-Rays from the sample or the shield material
Some of the energy from EM radiation will excite the electron in the shield material, will have prominent peak at the Xray energies of Lead
Xrays
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Xrays
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Energy lost from the photo peak which correspond to the energy equivalent of an electron
Only in high energy EM radiation (>1.022MeV) will escape peaks be present
If have a small peak 511 keV lower than your main photopeak this is the energy of an electron escaping from the detector
Single/Double Escape peak
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511 keV is a special energy It is the energy that is emitted when a
positron gets annilated by an electron and 2 511 keV gamma rays are produced.
If you get this peak in your spectrum there is a good chance you have a positron emitter in your sample
Easy to spot
511 keV
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Peaks
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If you have sample with much activity May get a peak that is twice the energy of
the photo peak This is from the detector seeing 2
independent EM radiations as one large one and will add the energies of them together and have one count but at twice the energy
Sum Peak
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Sum Peak
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Ge x-rays gets subtracted from the photopeak
Only important in low energy photopeaks In the order of 10-14 keV
Ge Escape Peak
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Full width half maximum Measure of the thickness of a peak Take the width in channels at the point
where the counts in those channels are ½ those of the maximum counts
The lower the number the more compact and peaky the photo peak will be
Can be used to determine if the detector is working correctly
FWHM
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Small detector-◦ size of detector is small compared to the path of
secondary gamma rays◦ Will result in significant compton continuum and
possible escape peaks Large detector-
◦ Detector is large compared to path of secondary gamma rays
◦ Will result in a single photo peak and no Compton continuum
Most detectors are in between
Detector Sizes
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Detector-detects radiation and produces an electronic signal
Amplifier- amplifies the signal so it can be better seen
ADC or Shaper- turns signal from Gaussian function into step function◦ 2 types used in counting
Scaler- separates signals according to potential
Counter- accumulates counts and displays results
Signal processing
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Energy-match energy of photo peak to the energy on the screen
Efficiency- match area of peak to activity of source
Calibration
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Purpose- match up the energy of the photopeaks with the energy label on the spectrum
How is it done- ◦ have a know spectrum of photo peaks and their
energies◦ Correlate the channel number to the photopeak
energy◦ Need to have at least 3 peaks(5 better)◦ Create an energy curve◦ Save data and apply it to all spectra
Energy Calibration
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Purpose- to have the area of the photopeak match the activity of the source
How is it done- ◦ have a standard for which you know the activity and
the % error◦ Assign that activity to the area of the photo peak◦ Need to have at least 5 peaks from low to high
energies◦ Create a efficiency curve that will be applied to all
spectra Efficiency will then be used to convert areas
to activities for all energies
Efficiency Calibration
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Low energies will have low efficiencies◦ EM can not effectively get through the protective
covering of the detector High energies will have low efficiencies
◦ Higher energy EM will have a lower probability of interacting with the detector
Efficiency Calibtration
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High energy beta will create brehmstrahhlung
This will appear on the gamma spectrum as a large broad low energy mound
Can interfere with small low energy peaks
Betas
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Neutron- too many neutrons will damage the detector
Avoid neutrons
Neutrons
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QUESTIONS?