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Physics 13 Laboratory

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Physics 13 Laboratory

    Physics 160

    Principles of Modern Physics

     Laboratory session #9

    Gamma Spectroscopy: Pre-Lab

    Answer the following questions PRIOR to coming to your lab section. You will not be allowed to participate in any data-collection until you have shown me your pre-lab and I have

    initialed it. Please tape or staple the pre-lab on the page opposite to the first page of your write-

    up; failure to do so will result in losing two points (out of a possible 20). Show all your work.

1. What is the main distinction between the Geiger-Müller tube you used in the last two labs and

    the scintillation detector you will use in this lab?

    2. What is the natural linewidth (in MeV) of the 0.662 MeV gamma ray emitted by Ba-137? Hint: You know that the half-life of Ba-137 is 153 s from last week’s lab. You will need to use

    the energy-time uncertainty principle.

3. Consider the interaction between a 0.662 MeV gamma ray and an electron in your scintillation

    detector. What is the maximum energy that can be transferred to the electron (in MeV) from the gamma ray via Compton scattering? Hint: You will need the Compton equation.

Physics 160 Laboratory

    Session 9: Gamma Spectroscopy 2

    Physics 160

    Principles of Modern Physics

    Laboratory session #9

    Gamma Spectroscopy

Objectives:

    ? To become familiar with the detection of gamma rays using a scintillation-

    photomultiplier tube detector and with the pulse height analysis technique for

    determining gamma ray energies. To understand the origin and location of the Compton

    edge and the backscatter peak in the pulse height spectrum.

    ? To calibrate the energy scale of the pulse height analyzer and use that calibration to

    measure the gamma ray energies of a number of other samples.

Introduction:

    So far this term we have used two different kinds of spectroscopy. We used photon emission

    spectroscopy to observe the energy levels of the hydrogen atom and electron-impact

    spectroscopy to observe those of the helium atom. We have seen that atoms emit discrete spectra

    due to the quantized nature of their energy levels, and that different atoms display different

    spectra. Quantum mechanics predicts that the energy states of nuclei are also quantized, and that

    spontaneous emission from nuclei of a particular radioactive isotope should therefore exhibit a

    characteristic spectrum of discrete energies. When photons are emitted in this process they are

    called gamma rays, and the technique of measuring and characterizing the discrete energy

    spectrum of gamma rays is known as gamma spectroscopy. While in essence similar to our

    previous spectroscopic experiments, gamma spectroscopy involves much higher energies, which

    require specialized apparatus for their detection. This week we will become familiar with the use

    of a scintillation detector and pulse height (or multichannel) analyzer for use as a gamma ray

    spectrometer.

Equipment:

    The principle component in the scintillation detector is a sodium iodide crystal (NaI). When a

    gamma ray from a radioactive sample enters the crystal, some combination of three physical

    processes can occur: 1) photoelectric emission of an electron that absorbs all of the gamma’s

    energy, 2) Compton scattering of the gamma ray photon off electrons in the crystal. or 3) pair-

    production of an electronpositron pair. In order for the last process to occur with any likelihood,

    the incoming gamma must have an energy that is at least twice the rest mass energy of the

    electron (2 X 0.511 MeV = 1.022 MeV). Although a couple of the samples you will use today

    emit gammas in this range, unless the gamma is substantially more energetic than 1.022 MeV,

    the pair-production mechanism is not observable. The electron liberated by the photoelectric

    effect is quite likely to scatter around in the NaI crystal, losing energy, until it is captured by an

    atom in the crystal with an electron vacancy. In the process of scattering, photons in the visible

    and UV region of the spectrum are emitted. Likewise with the Compton scattering process, the

    recoil electron will ultimately deliver most of its energy as visible and UV photons. The

    difference between the photoelectric and the Compton scattering process is that the former

Physics 160 Laboratory

    Session 9: Gamma Spectroscopy 3

process is likely to deposit all or nearly all of the incoming gamma energy it the crystal, while in

    the latter process, the scattered gamma ray photon may escape the scintillator crystal and

    therefore deposit only a fraction of its total energy in the crystal. The low frequency (visible

    and UV) photons produced when a gamma interacts with the scintillator crystal, enter a

    photomultiplier tube (PMT), in which a cascade of electrons is generated, again via the

    photoelectric (and secondary electron) effect. This has the effect of turning a light pulse into a

    current pulse, which is then converted into a voltage pulse. In general, the more energy the

    original gamma ray had, the larger the voltage pulse that the PMT will produce. The pulse

    height analyzer (PHA) divides the range of all possible voltages into bins, or channels, and keeps

    a running count of how many pulses arrive in each bin, thus producing a histogram of the

    number of counts versus PMT output voltage. Unfortunately, while the PMT voltage varies

    directly with gamma ray energy, that variation is not a simple proportion and it may not even be

    linear. This means that the scintillation detector must be calibrated with gamma rays of a

    number of known energies before it can be used to measure the energy spectrum of an unknown

    sample. The calibration results in a relationship that allows you to associate a given channel

    number with its appropriate energy.

Energy Calibration:

    Eight gamma ray emitting samples are available for investigation today. We will use three of

    them to calibrate the energy scale on the pulse height analyzer. In order to calibrate the energy

    scale over the full range of gamma ray energies we expect to observe, we will use Cd-109, Mn-

    54, and Co-60 as our calibration sources. Cobalt-60 has the highest energy gammas (1.173 MeV

    and 1.333 MeV) and you should start with this sample.

    ? Turn on the power switch for the SPECTECH Universal Computer Spectrometer box.

    ? Launch the UCS20 software from the Programs menu on the PC.

    ? From the SETTINGS menu, select HIGH VOLTAGE/AMPLIFIER.

    ? Turn detector voltage ON, and set voltage level to 500 V.

    ? Select the COARSE GAIN = 8.

    ? Place the Co-60 sample in the sample tray and place it on shelf #3.

    ? Start data acquisition.

    ? You should see a pulse height spectrum. Adjust the COARSE GAIN and the FINE

    GAIN until you obtain a spectrum for Co-60 that has the two prominent peaks near the

    far right end of the spectrum. These are the “photopeaks” associated with the

    photoelectric effect detection process discussed in the introduction for the 1.173 and

    1.333 MeV gammas. Once you find the right gain settings, do not alter them for the rest

    of the experiment.

    ? Acquire a good spectrum for Co-60 and identify the channel numbers that are at the

    center of each prominent peak. Note that the software has feature to help you determine

    the center of a peak. Feel free to explore these features, although it is satisfactory for the

    purposes of this lab to “eyeball” the center of the peak by moving the cursor across the

    peak. You will want to either print the spectra you obtain today or produce decent

    sketches of each spectrum, with labeled axes in your notebook. Next place the Mn-54

    sample in the sample tray and acquire a spectrum. Do NOT change the gain settings.

    Identify the channel number at the center of the peak corresponding to the 0.835 MeV

    gamma ray emission from Mn-54.

Physics 160 Laboratory

    Session 9: Gamma Spectroscopy 4

    ? Finally, acquire a spectrum for Cd-109 and identify the channel number associates with

    its 0.088 MeV gamma line.

    ? Use KALEIDAGRAPH to plot the known gamma energies versus the channel number at

    which each photopeak was centered. Does the relationship appear linear? Try a linear

    least squares fit. You should obtain an equation that relates the channel number to the

    energy deposited in the scintillator crystal. Note that this equation depends on the

    voltage applied to the PMT (500 V) and the amplifier gain. Altering either of these

    settings will alter the energy calibration equation.

    Source Gamma Energies of interest (MeV) Principle Decay mode Barium (Ba) 133 0.081, 0.276, 0.303, 0.356, 0.384 Electron capture Cadmium (Cd) 109 0.088 Electron capture Cesium (Cs) 137 0.662 Negative Beta Cobalt (Co) 57 0.122, 0.136 Electron capture Cobalt (Co) 60 1.173, 1.333 Negative Beta Manganese (Mn) 54 0.835 Electron capture

    Table I: Standard gamma sources

    Measurements:

    ? Check your calibration by acquiring spectra for the remaining samples in the table above,

    determining the energies of the characteristic gammas using your energy calibration

    equation, and comparing the energies obtained to the values given in the table above.

    ? Acquire the spectrum for Na-22. You should observe two prominent peaks. Determine

    the energies of each. The higher energy peak is associated with a transition from an

    excited state to the ground state in Ne-22, the daughter nucleus produced by the positive

    beta decay of Na-22. Positive beta decay involves the emission of a positron (or anti-

    electron). The positron eventually finds an electron and the two annihilate to produce

    two gamma ray photons that emerge at 180? to one another (to conserve momentum) and

    each has an energy equal to the rest mass energy of the electron (or positron). Only one

    of these two “annihilation” gammas can enter your detector. Verify that the energy

    associated with the lower energy peak in the spectrum of Na-22 has energy about equal to

    the rest mass energy of the electron.

    ? Identify the unknown: Place the unknown source under your detector. Acquire a

    spectrum of its gamma rays and identify its elements. Hint: the unknown contains one

    the calibration sources plus one of the following:

Physics 160 Laboratory

    Session 9: Gamma Spectroscopy 5

    Source Gamma Energies (MeV) Principle Decay Mode Sodium (Na) 24 1.368, 2.754 Negative Beta Zinc (Zn) 65 .511, 1.115 Positive Beta/Electron capture Silver (Ag) 108 0.433, 0.614, 0.723 Electron capture/Neg. Beta Rhodium (Rh) 102 0.475, 0.631, 0.698, 0.766 Electron capture/Neg. Beta

    Table II: Possible components of the unknown source.

Analysis and Questions:

    ? Return to the spectrum of Cs-137. Consider whether the width of the peak you measure

    is determined by the resolution of your instrument or whether you are seeing the true

    “uncertainty” or “natural linewidth” in the excited state of Ba-137, as determined in the

    prelab.

    ? Now consider the other features of the spectrum of Cs-137. Note that the channels just

    below the 0.662 MeV photopeak have very few counts, but that the “continuum” counts

    display a “shelf” somewhat below the energy of the photopeak. This continuum is called

    the Compton scattering continuum and it results from the Compton scattering of the

    incoming gamma off an electron in the crystal, and the subsequent escape of the gamma

    from the crystal after having lost some energy in the scattering process. Only the energy

    that the gamma loses inside the crystal is detected. The remaining energy in the photon is

    carried out of the detector and is not measured. Does the Compton “shelf” appear at the

    energy predicted by your prelab? There may be another peak on the Compton scattering

    continuum called the backscatter peak that coincides with a Compton scattering event in

    which the electron escapes and the scattered gamma ray photon is absorbed. Can you

    identify this peak? If so, at what energy does the backscatter peak occur and does this

    make sense?

    ? Use the Chart of the Nuclides to identify the decay scheme for all the samples you used

    today. In particular, identify the isotope associates with the nuclear transition that

    produced each gamma ray line you observed. For example, the 0.662 MeV gamma from

    Cs-137 is actually due to a transition in Ba-137, the daughter nucleus produced in the

    beta-decay of Cs-137.

Conclusions:

    Along with your usual sorts of conclusions, include such considerations as: How reliable is your

    calibration curve? What sort of uncertainty would be reasonable in your energies? What were

    the elements in the unknown sample? How does your data support this conclusion?

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