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

Gamma SpectroscopyNUCP 2371

Radiation Measurements II

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

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

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

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

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

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)

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

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

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

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

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

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

NaI and Ge detectors

Uranium Spectrum

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

Spectrum

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

Photopeak

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

Xrays

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

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

Peaks

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

Sum Peak

Ge x-rays gets subtracted from the photopeak

Only important in low energy photopeaks In the order of 10-14 keV

Ge Escape Peak

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

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

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

Energy-match energy of photo peak to the energy on the screen

Efficiency- match area of peak to activity of source

Calibration

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

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

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

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

Neutron- too many neutrons will damage the detector

Avoid neutrons

Neutrons

QUESTIONS?