ASSIGNMENT OF INSTRUMENTAL CHEMISTRY

TOPIC:- GAMMA-RAY SPECTROSCOPY

Submitted by =Moin Khan Hussain Submitted to= Ms.Hardeep kaur

UID-14BBT1129Course-Bsc. Biotech

Abstract:This experiment was mainly to observe the spectra of different sources, to understand the nature of what it represents. To calibrate the software and work out the efficiency of the PMT.n, the number of dynodes was found to be 6.5R the resolution of the spectrometer was found to be 0.130 +/- 0.006R in part two was found to be inversely proportional to the energy of the gamma ray.The amount of nuclei was estimated to be compared with the empirical result of:

.

Introduction:

Gamma-ray spectroscopy is based on one vitally important instrument. The photomultiplier tube (PMT). The PMT works by (see figure 1) allowing an incoming photon to be absorbed by its photocathode. The then produced photoelectron is then accelerated by the voltage and is bounced off a number of dynodes. Each which “multiply” the current travelling through the tube by a mean factor m. This factor is typically, for a few hundred volts of the order of about 4. The spectrum read from this photocathode is of the number of pulses (counts) vs. Pulse height (Energy). The energy is proportional to the energy of the incoming photon, as this is what gives the photoelectron its initial momentum. But the energy is also very dependent on the voltage across the photomultiplier tube. Thus any x-axis values must be calibrated first and the voltage kept constant.

Where do the photons come from? Well it’s quite roundabout. In summary the gamma rays transfer their (~1MeV) energy to electrons in the material surrounding the source, which in turn transfer their energy to thousands of (~3eV) longer photons via a scintillator covering the PMT.The gamma rays transfer their energy to the electrons in a number of ways:

Photelectric absorption: The most useful for determining the gamma ray energy. The gamma ray gives all of its energy into expelling an electron from one of the inner-most shells of an atom which then has an energy and leaves a hole behind. When this hole is filled up an x-ray photon is released which itself often undergoes photoelectric absorption giving both electrons a total energy of . Hence why we call this peak the “total energy peak”

Compton Scattering: The gamma ray photon here is deflected by an electron. The scattered photon now has less energy than the incident and thus of a longer wavelength. The electron receives the energy lost by the photon. By equating the principal of conservation of momentum with relativistic conservation of energy, Compton scattering wavelengths can be expressed as:

Which is the energy expression. The max energy a scattered photon can have is:

Pair Production: Should the gamma ray exceed twice the rest mass energy of the electron (ie 2*0.551MeV). An electron positron pair may be created in the vicinity of the nucleus. The positron will soon annihilate with a nearby electron releasing two gamma ray photons (each of 0.551MeV) in opposite directions. The reason there are two is because of conservation of momentum.

This process gives a distinctive spectrum (see appendices 1, 2, and 3 for examples) containing a back-scatter peak, a Compton edge, a total energy peak and/or a sum peak.The resolution of the spectrum, ie the reason a sharp peak is not present is as a result of two reasons: a) the statistical nature of the process by which the gamma ray photon gives rise to a charged pulse and b) the imperfect nature of the apparatus. It is usual to quote the resolution as a ratio of its full width at half height to the pulse height of the peak.

Experimental Details:

(i)The aim of this experiment was to record and analyse the effect that changing the photomultiplier voltage had on the spectrum.

The apparatus was set up as in Fig. 2 above. The FWHM (full width at half height of max) was taken as well as the relative pulse height (ie channel no.) of the max and plotted against varying voltage..

(ii)The aims of this experiment were to:-Record and interpret the spectra for Cs137, Co60 (with s at 1.137 and 1.332 MeV) and Na22-Investigate the dependence of resolution, R, on -ray energy.

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The software was first calibrated then the spectrum was taken for the above three sources. Each peak was noted and compared with known values. R (the resolution) was measured for each total peak.

(iii)The aims of this experiment were to:-Measure the -ray spectrometer’s detection efficiency..-To measure the activity of a sample of Potassium Chloride, which is in low-sodium salt.

A sample of KCl was placed at the aperture for a few days to let a significant count build up. The efficiency of the apparatus for different energies was measured and used to calculate the count rate of the potassium 40 in the sample.

Results and Analysis:

(i) Why?

Well since each dynode multiplies the current by a mean factor of then the overall gain from all the dynodes will be , which is proportional to the pulse height (for charged particle of fixed energy). The factor is proportional to the mean voltage across each dynode: which is proportional to the total voltage across all the dynodes:

. Therefore: , the pulse height is proportional to the voltage to the power of the number of dynodes.

Results were taken of H0 vs. V (see fig 2). The results obviously weren’t linear so n was increased until the best fit straight line graph was found. n=6 and n=7 were found to be equally but oppositely off the mark.

n=6.5 gave the best fit line (see fig 3).It seems an odd considering n would

be expected to be a natural number. The extra half may be from the contribution of the smaller anode to multiplication. Or the calibration may have

become offset midway through the experiment.Whether these have any truth is uncertain. The fact remains however that this number best fits the results obtained and thus shall be taken to be the case for the remainder of the experiment.

Fig 3

Fig 4

How does FWHM and R depend on V?Since R is merely a ratio of Gaussian distribution it should remain constant with varying voltage. The results giveSince FWHM divided by gives R, a constant, then FWHM acts similarly to with varying voltage

Since R is relatively constant for our measurements of V, the value should be the average of all the values of R with a standard error in the mean:R=0.130 +/- 0.006The value obtained for V=0.9kV was: R=0.131 which agrees well within experimental error.

(ii)After calibration the following features in the spectra were observed:

(see appendix 2)Backscatter peak: 0.29MeVX-ray peak: 0.02MeVCompton Edge: 0.52MeVTotal Energy Peak: 0.67MeV

From the book Knoll, the decay process given for gives 0.662MeV as the energy of this gamma ray, corresponding with the total energy peak.

Backscatter peak: 0.17MeVX-ray peak: 0.07MeVCompton Edge: 0.40MeVTotal Energy Peak: 0.51MeV

The total energy peak above is characteristic with the two 0.511MeV gamma rays released when the beta positron annihilates with an electron. There is another peak at 1.274 (not apparent in the spectrum at the time) caused from the initial release of a gamma ray.

(see appendix 3)Backscatter peak: 0.15MeVCompton Edge: 0.99MeV

: 1.17MeV: 1.33MeV

Sum Peak total: 2.5 MeV

This decays with emission of a beta- particle and two cascading gamma rays of 1.17MeV and 1.33MeV, exactly as found.

As V is remaining constant, the only other variable that R is dependent on is E; the energy of the incident gamma ray. R is inversely proportional to E so when multiplied, in theory they should equal a constant.The following values of R, E and thus K (constant) were found for the and :

R E (MeV) K0.22 0.51 0.110.15 0.67 0.10

The constants above are, for all intents and purposes, the same. This shows that the result for R matches the expected value.

R for the could not be found in the usual way due to the presence of a “sum peak”. So taking the energy as of the peaks and using the mean value of the constants; R for worked out to be 0.04.

(iii)The total energy peak for (at 0.551MeV) had a relative number of pulses (count rate) of 1494The sum peak relative number of pulses were 200 and 184.

Thus

Similar calculations for of at an approx ~1.2MeV energy. Finding

(see Appendix 1 for spectrum of )

To work out the apparent nuclei, the activity must be worked out by taking the count rate divided by the efficiency. But to get the count rate for this we must integrate across the peak itself and subtract the background radiation (see Fig 4). Thus the shape of the graph was assumed to be:

(eq. 1)Therefore

Where the second integral was worked out from the background line and the limits were calculated from equation 1. For convenience the max extremum was taken to be at x=0.

Fig 5

Only 11% of the nuclei will emit a gamma ray because of its decay branch ratio. So the corrected value is:

This corresponds within an order of magnitude of the previously calculated estimation. This is all that can be expected considering the inaccuracy of this method.

Conclusions:n, the number of dynodes was found to be 6.5. Which is probably incorrect. Yet the results best fit 6.5, of that there is no doubt. As stated before this may be as a result of the smaller anode’s contribution to multiplication, this is just speculation though.R the resolution of the spectrometer was found to be 0.130 +/- 0.006

R in part two was found to be inversely proportional to the energy of the gamma ray when the PMT voltage was kept at a constant.

The efficiency of the spectrometer for gamma-ray photons of 0.551MeV was found to be:

For 1.2 MeV it was found to be: The amount of nuclei was estimated to be . Using the spectrometer it was worked out to be . This is within an order of magnitude which is (as expected) as accurate as possible considering the imprecise nature of the method.

References:-

https://en.wikipedia.org/wiki/Gamma_spectroscopy http://www.ortec-online.com/Solutions/gamma-

spectroscopy.aspx http://www.sciencedirect.com/science/article/pii/

0029554X76904729