Gamma-ray Spectroscopy
Declan Cockburn: 02038633.
Lab partner: William Walshe.
February 2005
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 6.5
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.
K 40 nuclei was estimated to be 3.24  1018 compared with the empirical result of:
5 1017 nuclei .
The amount 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 Eb and leaves a hole behind. When this hole is filled up an xray photon is released which itself often undergoes photoelectric absorption giving both
electrons a total energy of E . 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:
 '  
 h ' 
h
1
1 (1  Cos ) 1
h (1  Cos )  me c 2
(1  Cos ),


,

me c
h ' h
me c 2
h '
me c 2h
me c 2h
h

Q.E.D.
2
h (1  Cos )
h (1  Cos )  me c
[1 
]
me c 2
Which is the energy expression. The max energy a scattered photon can have is:

h
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 backscatter 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.
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)
H 0  V n Why?
Well since each dynode multiplies the current by a mean factor of
m then the overall gain from all the
n
dynodes will be m , which is
proportional to the pulse height
(for charged particle of fixed
energy). The factor m is
proportional to the mean voltage
across each dynode: Vd which
1000
800
H0
is proportional to the total
voltage across all the dynodes:
1200
600
V . Therefore: H 0  V n , the
400
pulse height is proportional to
the voltage to the power of the
number of dynodes.
200
0
0
0.2
0.4
0.6
kV
1
Fig 3
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.
1200
1000
800
H0
0.8
600
n=6.5 gave the best fit line (see fig 3).
It seems an odd considering n would
400
be expected to be a natural number.
The extra half may be from the
200
contribution of the smaller anode to
multiplication. Or the calibration may
0
have become offset midway through
0
0.2
0.4
0.6
0.8
1
1.2
the experiment.
Fig 4
kV^n
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.
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 give
Since FWHM divided by H 0 gives R, a constant, then FWHM acts similarly to H 0 with varying
voltage  FWHM  V
n
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:
1.2
R=0.130 +/- 0.006
The 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:
Cs137 (see appendix 2)
Backscatter peak:
X-ray peak:
Compton Edge:
Total Energy Peak:
0.29MeV
0.02MeV
0.52MeV
0.67MeV
137
From the book Knoll, the decay process given for Cs
ray, corresponding with the total energy peak.
gives 0.662MeV as the energy of this gamma
Na 22
Backscatter peak:
X-ray peak:
Compton Edge:
Total Energy Peak:
0.17MeV
0.07MeV
0.40MeV
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.
Co60 (see appendix 3)
Backscatter peak:
Compton Edge:
E1 :
0.15MeV
0.99MeV
1.17MeV
E2 :
1.33MeV
2.5 MeV
Sum Peak total:
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.
22
137
The following values of R, E and thus K (constant) were found for the Na and Cs
R E (MeV)
K
Na 22
Cs137
0.22
0.51
0.11
0.15
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.
60
R for the Co could not be found in the usual way due to the presence of a “sum peak”. So taking the
energy as
be 0.04.
E1  E2 of the peaks and using the mean value of the constants; R for Co60 worked out to
(iii)
22
The total energy peak for Na (at 0.551MeV) had a relative number of pulses (count rate) of 1494
The sum peak relative number of pulses were 200 and 184.
Thus
2 A  1494
A 2  200  184  384
 4 A 2  2 A
2 2    0
1
 0.551 
2
Similar calculations for
Finding
1.2  0.013
 of Co60 at an approx ~1.2MeV energy.
K 40
3.625g KCl  0.049mol
 K 40  5.4  10 6 mol  3.24  1018 radioactive nuclei
 abundance  0.011%
(see Appendix 1 for spectrum of
K 40 )
To work out the apparent nuclei, the activity must be
worked out by taking the count rate divided by the
40
efficiency. But to get the count rate for this K 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:
y   x 2  550 (eq. 1)
Therefore
23.5
 ( x
 23.5
23.5
2
 550)dx 
 (5x  220)dx
 23.5
 total counts  3000
Fig 5
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.
A  count rate  1.23  102 s 1
1.39  1.2  0.013
dN (t )
 N ( t )
dt
T1  (0.94 s 1 )
1
0.94 s
 N (t ) 
 2

ln 2
16
1
4  10  (0.94 s )

 5.5  1016 apparent nuclei
ln 2
 A  0.94 s 1 
Only 11% of the nuclei will emit a gamma ray because of its decay branch ratio. So the corrected
value is:
 K 40  5 1017 nuclei
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:
1.2  0.013
 0.551 
1
2
K 40 nuclei was estimated to be 3.24  1018 . Using the spectrometer it was worked
17
out to be 5 10 nuclei . This is within an order of magnitude which is (as expected) as accurate as
The amount of
possible considering the imprecise nature of the method.