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NUCP 2371 Radiation Measurements II. Since electromagnetic radiation does not cause any direct ionization Detector system needs to do several things Reasonable.

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Presentation on theme: "NUCP 2371 Radiation Measurements II. Since electromagnetic radiation does not cause any direct ionization Detector system needs to do several things Reasonable."— Presentation transcript:

1 NUCP 2371 Radiation Measurements II

2 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

3 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

4 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

5 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


7 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

8 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


10 λ 1 - λ 0= h/m e c (1- cosθ) λ 1 = wave after scattering λ 0 = initial wave of EM radiation h/m e c= Compton wavelength of electron 2.43 X 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

11 Determine λ of initial EM radiation Calculate change in λ Add change in λ to original λ Then calculate the new energy of the scatter EM radiation Start with 500 keV photon that is deflected 45degrees What next?

12 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



15 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

16 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

17 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



20 Photo peak Xrays Compton edge Single escape peak Double escape peak Sum peak 511 peak Ge escape peak FWHM

21 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


23 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


25 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

26 511 keV is a special energy It is the energy that is emitted when a positron gets annilated by an electron and 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


28 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


30 Ge x-rays gets subtracted from the photopeak Only important in low energy photopeaks In the order of keV

31 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

32 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

33 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

34 Energy-match energy of photo peak to the energy on the screen Efficiency- match area of peak to activity of source

35 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

36 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

37 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

38 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

39 Neutron- too many neutrons will damage the detector Avoid neutrons


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