Radiation Measurements
M. Nomachi
July, 2012
1
Purpose
The gamma rays from natural radioactivity in
processing and Statistical analysis is studied.
40
2
40 K
is measured with CsI(Tl) detector. Signal
K
is one of natural radioactivities. It exists everywhere. Half-life of 40 K is 1.265 × 109 years.
Natural abundance of 40 K is 0.0117 %. The decay scheme is shown in Figure 1.
40 K
Figure 1:
2.1
40 K
decay scheme
advanced study
1. What is ”Elctron Capture”?
2. Why the life time of
40 K
is so long?
3. Why the branching rate to 40 Ar ground state is very much smaller than that to the excited
state?
2.2
Half-life
Radiative decay of unstable nuclei is a statistical event. The probability of disintegration per
unit time is constant. Assuming the decay probability µ, The number of decay is promotional
1
to the number of radioactive nuclei.
dN = −µN dt
(1)
−µt
(2)
N = N0 e
At the time T1/2 the number of nuclei becomes half. T1/2 is half-life.
1
N0 = N0 e−µT1/2
2
loge 2 = µT1/2
loge 2
µ=
T1/2
(3)
(4)
(5)
Half-life of 40 K is 1.265 × 109 years. Decay probability is loge 2/1.265 × 109 = 0.548 × 10−9 /year.
One year is 3.16 × 107 seconds. So, decay probability for one second is 1.74 × 10−17 /sec. One
mol (6.02 × 1023 ) of 40 K has 1.05 × 107 decay per second. Or, 1.05 × 107 Bq.
2.3
Potassium chloride KCl
We measure radioactivity of 40 K in salt. Normal salt is mainly NaCl and KCl concentrarion is
small. We use ”Potassium enriched salt” for the measurement. It is ”Yasashio” from Ajinomoto.
It has 27.6g of KCl in 100g. Mass number of Cl is 35.45 g/mol. Mass number of Potassium (K) is
39.10 g/mol. Therefore, the amount of potassium in 100g of the salt is 27.6/(35.45+39.10)=0.37
mol. Thus 40 K is 0.37 mol ×0.0117% = 4.3 × 10−5 mol. The decay persecond in 100g of salt is
450 Bq. 10.67% of decay emitts 1.46 MeV γ ray. It means about 50 γ rays/second/100g-salt.
3
Poisson distribution
Radioactive decay is statistical process. The probability is obtained but exact time cannot be
known. The probability of zero disintegration for the period T is
P0 (T ) = e−µT
(6)
The probability of just one disintegration is that summation of having disintegration at time t
to t + dt and no disintegration in the rest of the period. Thus,
∫ T
P1 (T ) =
e−µT µdt = µT e−µT
(7)
0
The probability of just two disintegrations is that summation of having disintegration at time t
to t + dt and one disintegration before and no disintegration after. Thus,
∫ T
∫ T
(µT )2 −µT
−µt −µT −t
e
(8)
P2 (T ) =
µte e
µdt =
µte−µT µdt =
2
0
0
and so on. This probability is known as Poisson distribution.
Pn (T ) =
(µT )n −µT
e
n!
(9)
To study Poisson distribution, we introduce the generating function.
G(x) =
∞
∑
n=0
n
Pn (T )x =
∞
∑
(µT x)n
n!
n=0
2
e−µT = eµT (x−1)
(10)
G(1) is normalized to 1.
G0 (x) =
∞
∑
nPn (T )xn−1 = µT eµT (x−1)
(11)
n=0
G0 (1) is average and it is µT .
The mean squar of the measured number of events in period T is defined as
∞
∑
(n − µT )2 Pn (T ) =
n=0
∞
∑
(n − G0 (1))2 Pn (T ) =
n=0
∞
∑
(n2 − 2nG0 (1) + G0 (1)2 )Pn (T )
(12)
n=0
It is calculated with the generating function
(∞
)
∞
∑
d ( 0 )
d ∑
n
xG (x) =
nPn (T )x
=
n2 Pn (T )xn−1
dx
dx
n=0
(13)
n=0
= G0 (x) + xG00 (x)
(14)
Therefore, mean square is
G0 (1) + G00 (1) − 2G0 (1)2 + G0 1(1)2 = G0 (1) + G00 (1) − G0 (1)2
(15)
= µT + (µT ) − (µT ) = µT
(16)
√
Consequently, we obtain
the root mean squar (rms) of µT . Measureing N events, the rms of
√
measured value is N .
2
4
2
Propagation of error
Often we calcurate a value from several variables with error. The error of the results is evaluated
from the error of variables. Let’s assume the function z = f (x, y). Error of x is ∆x. Error of y
is ∆y. Here, we assume ∆x and ∆y are independent each other. Fractuation δz is
δz =
∂f
∂f
δx +
δy
∂x
∂y
(17)
Since δx and δy are independent,
(
2
∆z =
In the case of z =
)2 (
)2
∂f
∂f
∆x +
∆y
∂x
∂y
x
,
y
(
∆z
(
5
(18)
∆z
z
2
)2
)
x∆y 2
=
+
y2
(
)2 (
)
∆x
∆y 2
=
+
x
y
∆x
y
)2
(
(19)
(20)
Experiment-1
The number of counts related to the source is measured by subtracting the counts without source
from the counts with source. Since the radioactivity in the source is so low that the difference
is very small. To obtain meaningful results, the number of hits associate to the 40 K should be
larger than the statistical fractiatuon of the number of the background hits.
3
5.1
5.1.1
Setup
CsI detector
CsI(Tl) is inorganic scintillator. [2] The size of scintillator is 1cm × 1cm × 1cm
Figure 2: CsI detector and amplifire / counter
The detector module has a Pre-amplifier, a Shaper amplifier and a Discriminator. The
number of discriminator output is counted. It needs -9V power supply to operate.
Figure 3: Front panel of the counter
Pressing RST & ST button resets counter and starts counting. In AUTO mode the counter
stops in predefined TIME. The TIME is set in the switch at the front panel. One needs to use
a small sqrew driver.
5.2
Measurements and Analysis
The radioactivity of the salt is very low. We must care about the error. Measurening N events
in time T , the counting rate of N/T is obtained. Since N statistically fractuates. The number
of events fracuate as Poisson distribution. The rate calculated from the measured value has
statistical error.
4
1. Estimate the event rate you may obtain with the source.
2. Repeat the measurments and get the distribution of the measured value.
3. Compare the distribution with Poisson distribution
4. (Advanced) Discuss the difference to the theory. Can you make the difference small? How?
The measured counts contains not only the signal but also the background. It is necessary
to measure in two setup, i.e. with the salt (source run) and without the salt (background run).
Subtracting the result of the background run from the resylt of the source run, one can obtain
the net counts of the salt,
1. Study how the error propagetes in the calcuration.
2. (Advanced) Optimize the time for the source run and the time for the background run.
5.3
Counting rate
Assuming uniform and infinite volume of Yasashio (KCl enriched salt), how many gamma rays
are absorbed in unit mass? Since 50 gamma rays/second are generated in 100g, 50 gamma
rays/second should be absorbed in 100g. Replacing a part of salt by CsI(Tl) detector, the same
ammount of gamma rays/second should be absorbed in CsI(Tl) detector. 1 cm3 of CsI(Tl)
detector is 4.51 g. Thus 2.3 gamma rays/second can be expected. It is too much simplified
calcuration but the order of counts should be not so different.
6
6.1
Experiment-2
USB Modular Oscilloscope
The Agilent U2701A oscilloscope is USB-based modular oscilloscope. [1] It is a 2 channel
100MHz, 8-bits oscilloscope. The output of the amplifier should be connected. The pulse shape
of the CsI detector is observed in the application provided by Agilent.
Figure 4: U2701 USB osciloscope
Through the windows on the PC, the Agilent Measurement Manager (AMM) software provides a front-panel interface.
5
Figure 5: A PC window for U2701 USB osciloscope
6.2
Measurement-1
Observe the wave form of the Amplifire output. The output is found at the rear panel of
the Preampifier / Counter module. The output cannot drive 50 Ω terminator. It should be
connected without a terminator.
1. Observe the width of the signal
2. Observe the typical pulse hight of the signal.
6.3
Agilent VEE
Agilent VEE is programing tool for data taking and analysis. It is similar to LabView software.
The waveform of the signal is measured. A simple example is provided. The example makes
a histogarm of the One can modify to improve the measurements. The example saves a list of
pulse height in a file. One can analyze the data after the measurement.
Figure 6: Energy spectrum
6
The figure shows the data with high statistics. The data used takes several hours. The peak
seen around 1500 is the peak of the gamma rays from 40 K.
1. (Advanced) Why the number of data in the peak is so small.
References
[1]
http://www.home.agilent.com/agilent/product.jspx?nid=-34492.0.00&cc=VN&lc=eng
[2]
http://www.detectors.saint-gobain.com/uploadedFiles/SGdetectors/
Documents/Product_Data_Sheets/CsI(Na)-CsI(Tl)-Data-Sheet.pdf
7