1) The experiment used gamma-ray spectroscopy to analyze spectra from various radioactive sources. Spectra were recorded at different photomultiplier tube voltages to study resolution and efficiency. 2) Analysis found the number of dynodes in the photomultiplier tube to be 6.5, and resolution R was determined to be inversely proportional to gamma ray energy as expected. 3) Activity of a potassium chloride sample was estimated using detector efficiency calculations, finding 1.7×10^17 40K nuclei, consistent with the expected amount.

1. ASSIGNMENT OF INSTRUMENTAL
CHEMISTRY
TOPIC:- GAMMA-RAY SPECTROSCOPY
Submitted by =Moin Khan Hussain
Submitted to= Ms.Hardeep kaur
UID-14BBT1129
Course-Bsc. Biotech

2. 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.
The amount of 40
K nuclei was estimated to be 18
1024.3 × compared with the empirical result of:
nuclei17
105× .
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 bE 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 γE . 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:

3. ...
]
)1(
1[)1(
'
)1(
'
1
,
)1(1
'
1
),1('
2
2
2
2
2
2
DEQ
cm
Cosh
h
cmCosh
hcm
h
hcm
cmCosh
hcm
Cos
hh
Cos
cm
h
e
e
e
e
e
ee
θυ
υ
θυ
υ
υ
υ
θυ
υ
θ
υυ
θλλ
−
+
=
+−
=∴
+−
=
−
=−−=−
Which is the energy expression. The max energy a scattered photon can have is: υh
 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 γs at 1.137 and 1.332 MeV) and Na22
-Investigate the dependence of resolution, R, on γ-ray energy.

4. 0
200
400
600
800
1000
1200
0 0.2 0.4 0.6 0.8 1 1.2
kV^n
H0
0
200
400
600
800
1000
1200
0 0.2 0.4 0.6 0.8 1 1.2
kV
H0
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 γ-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)
n
VH ∝0 Why?
Well since each dynode multiplies the current by a mean factor of m then the overall gain from all the
dynodes will be n
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: dV which
is proportional to the total
voltage across all the dynodes:
V . Therefore:
n
VH ∝0 , 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

5. 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 0H gives R, a constant, then FWHM acts similarly to 0H with varying
voltage n
VFWHM ∝∴
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 +/- 0.006
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:
137
Cs (see appendix 2)
Backscatter peak: 0.29MeV
X-ray peak: 0.02MeV
Compton Edge: 0.52MeV
Total Energy Peak: 0.67MeV
From the book Knoll, the decay process given for 137
Cs gives 0.662MeV as the energy of this gamma
ray, corresponding with the total energy peak.
Backscatter peak: 0.17MeV
X-ray peak: 0.07MeV
Compton Edge: 0.40MeV
Total 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.
60
Co (see appendix 3)
Backscatter peak: 0.15MeV
Compton Edge: 0.99MeV
1E : 1.17MeV
2E : 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 22
Na and 137
Cs :
22
Na

6. R E (MeV) K
22
Na 0.22 0.51 0.11
137
Cs 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.
R for the 60
Co could not be found in the usual way due to the presence of a “sum peak”. So taking the
energy as 21 EE + of the peaks and using the mean value of the constants; R for 60
Co worked out to
be 0.04.
(iii)
The total energy peak for 22
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
14942 =εA
2
1
02
24
384184200
551.0
2
2
2
≈
≈−
≈∴
=+=
ε
εε
εε
ε
AA
A
Similar calculations for ε of 60
Co at an approx ~1.2MeV energy.
Finding 013.02.1 ≈ε
%011.0
1024.3104.5
049.0625.3
18640
40
=
×=×=∴
=
−
abundance
nucleieradioactivmolK
molKClg
K

(see Appendix 1 for spectrum of 40
K )
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 40
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:
5502
+−= xy (eq. 1)
Therefore
3000
)2205()550(
5.23
5.23
5.23
5.23
2
==
+−+− ∫∫ −−
countstotal
dxxdxx
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

7. nucleiapparent
s
sT
s
tN
tN
dt
tdN
sA
sratecountA
16
116
1
2
11
1
2.139.1
12
105.5
2ln
)94.0(104
2ln
)94.0(
94.0
)(
)(
)(
94.0
013.0
1023.1
×=
××
=
×
==∴
===∴
≈≈
×==
−
−
−
−
−−
λ
λ
εε
ε
Only 11% of the nuclei will emit a gamma ray because of its decay branch ratio. So the corrected
value is:
nucleiK 1740
105×=∴
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:
2
1
551.0 ≈ε
For 1.2 MeV it was found to be: 013.02.1 ≈ε
The amount of 40
K nuclei was estimated to be 18
1024.3 × . Using the spectrometer it was worked
out to be nuclei17
105× . This is within an order of magnitude which is (as expected) as accurate as
possible considering the imprecise nature of the method.
References:-

8.  https://en.wikipedia.org/wiki/Gamma_spectroscopy
 http://www.ortec-online.com/Solutions/gamma-
spectroscopy.aspx
 http://www.sciencedirect.com/science/article/pii/0029554
X76904729