experiments with the alpha/betadetectorlavfis2/bancoapostilasimagens/apradgama... · 2010. 3....

30.0 y

0.662

0

56Ba137

55Cs

137

93.5%

6.5%

+ -

U

R

Signal

Guldfolium

Skikt av p-typ

Utarmnings-område

Si-kristallav n-typ

Laddad partikel

Experiments with theALPHA/BETAdetector

Teacher´s Handbook & StudentInstruction

P.O. Box 15120, SE-750 15 UPPSALA, SWEDENPhone: +46 18 480 58 00, Fax: +46 18 555 888

E-mail: [email protected], Internet: www.gammadata.net

User’s Guide for he Alpha/ & Beta Detector version 3.2

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Teacher’s Handbook

for the Alpha & Beta Detector

Teacher’s Handbook



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1. Investigation of alpha and beta radiation 1

1.1 Purpose 1

1.2 Time needed 1

1.3 Required theoretical knowledge 1

2. Experimental procedure: alpha radiation 2

3. Experimental procedure: beta radiation 73.1 Formulas for calculating Q values for β decay 7

© 1997 Gammadata M�tteknik AB

Contents Page


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To illustrate the properties of alpha and beta radiation. The energy loss of alpha particlesin air and aluminium is studied. The energy difference between the K and L atomicshells of the daughter nuclide is measured with the aid of conversion electrons.

1 - 2 hours per lab exercise.

Basic knowledge of the components of atomic nuclei. Use of the Chart of Nuclides.

1

1. Investigation of alpha and beta radiation

1.1 Purpose

1.2 Time needed

1.3 Required theoretical knowledge


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The experiment is performed with a GDM system equipped with the optional alpha/betadetector, and alpha and beta radiation sources mounted on rods. The alpha/beta detector was designed for Risø's as well as Gammadata's alpha and beta sources. The Risø alpha source has, however, one major drawback: It is encapsulated by a thincoating, which decreases the energy of the particles. The energy distribution is alsobroadened because of the different paths the particles can follow through the coating.Figure 1 compares this distribution with one obtained from an unshielded source. Theaverage energy loss of the alpha particles from the 241Am Risø source is about 0.8 MeV.This must be considered in the calculations in Experiment 1. The corresponding energyloss in Gammadata’s sources is negligible.

Figure 1. Spectrum from the Risø alpha source (1) and from Gammadata’s alpha source (2)

2

100 200 300 400 Channels

200

600

1000

1400

Counts

1

2

2. Experimental procedure: alpha radiation


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The alpha experiments give the best results with a source with very thin coating. Gammadata manufacture such a source.

Figure 2. Results from Experiment 1 using Gammadata's alpha source

Figure 2 shows the spectrum in Experiment 1. The rightmost peak was obtained with thesource rod fully inserted, the other peaks with the rod pulled out by 0.5, 1.0, 1.5, 2.0, and2.5 cm, respectively. The distances should be measured accurately, e.g. with simplegauge blocks. The gain is adjusted with the trim potentiometer so as to put the peak atabout channel 480 with the rod fully inserted.

3

Counts

Channels

200

150

100

50

100 200 300 400


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Diagrams 1 and 2 in the Appendix give the energy loss for charged particles in differentmaterials. The diagrams refer to protons, but can be converted for other charged par-ticles by means of the formulas provided.

Diagram A shows what Diagram 1 in Experiment 1 can look like. Notice the two energyaxes. First, the axis is scaled in ”channels”. Once the curve has been sketched and ex-trapolated to the energy axis, it is scaled in MeV.

Figure 3. Diagram A

4

4

1 2 3 4

-energy

Distance

cm

500

400

300

200

100

Channels

-energyMeV

5

3

2

1

5,48

α

α


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Diagram B shows the stopping power dE/dx as a function of the distance betweensource and detector (Diagram 2 in Experiment 1). The stopping power was obtained as(the magnitude of) the slope of the graph in Diagram A. The increase at the right-handend is known as the ”Bragg peak”.

Figure 4. Diagram B

5

1 2 3 4

Stopping power

Distance

cm

4

3

2

1

MeV/cm


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Figure 5. Results from Experiment 2 using Gammadata's alpha source.

Figure 5 shows a spectrum from Experiment 2 recorded with an unshielded source.

From Diagram A one finds that the energy of the alpha particles is reduced from about

4.9 MeV to 2.4 MeV. (The mean energy is then 3.6 MeV.)

Using Diagram 2 in the Appendix for Eα = 3.6 MeV (Ep = 0.9 MeV) one finds that dE/dx is

800 keV/(mg/cm2). The thickness of this aluminium foil is accordingly 3 mg/cm2. Given

the density of aluminium, 2.7 g/cm3, the thickness of the foil is found to be 0.01 mm.

6

100 200 300 400 Channels

Counts

200

400

600


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The Risø gamma-ray source (137Cs) could be used as described in the beta-ray exercise.However, this source also seems to be covered with a coating that stops electrons so that theconversion peaks cannot be seen in the spectrum. Risø’s beta-ray source contains 90Sr, which does not emit any conversion electrons. Nevertheless, it can be used to illustrate the appearance of a beta-ray spectrum.

Figure 6. Beta spectrum from 137Cs

The beta spectrum from 137Cs is shown in Figure 6. The conversion-electron peaks can be seen on the right-hand side of the spectrum. The larger peak is due to K electrons, while L electrons give rise to a small peak to the right of the K peak.

As was mentioned above, Risø’s cesium source cannot be used in this lab exercise. Gammadata Mätteknik AB have manufactured a source. (Permit issued by the Swedish Radiation Protection Institute, October 10, 1991, document no. 6541/11.) It serves as both a beta source and a gamma source.

Qβ- = mxc2 - myc2

Qβ+ = mxc2 - myc2 - 2mβc2

NOTE: mx and my are the atomic masses of the mother and daughter nuclide, respectively.

Qβ- = Eb(y) - Eb(x) + ∆mc2

Qβ+ = Eb(y) - Eb(x) - ∆mc2 - 2mec2

where: Eb(x) and Eb(y) is the binding energy of the mother and daughter nucleus, respectively.

∆mc2 = (mn - mp - me)c2 = 0.7824 MeV

mec2 = 0.511 MeV

7

3. Experimental procedure: beta radiation

0 0,50 1,00 1,50

MeV

1000

2000

3000

4000

5000

E

Counts

0

3.1 Formulas for calculating Q values for β decay

β


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Student Instruction


Student Instruction



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1. Investigation of alpha radiation 1

1.1 Purpose 1

1.2 Equipment 1

1.3 Theory 1

1.3.1 Alpha decay 1

1.3.2 The detector 2

1.4 Experiment – Stopping power of air for alpha particles 3

1.4.1 Diagram A – Stopping power of air for alpha particles 4

1.4.2 Diagram B – Stopping power of air for alpha particles 5

1.5 Experiment – Determination of the thickness of an aluminium foil 6

2. Investigation of beta radiation 7

2.1 Purpose 7

2.2 Equipment 7

2.3 Theory 7

2.3.1 Beta decay 7

2.3.2 The detector 8

2.4 Experiment – Study of beta particles and conversion electrons 9

3. Exercises 10

© 1997 Gammadata M�tteknik AB

Contents Page


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To investigate alpha radiation with the aid of a surface-barrier detector. The energy lossof alpha particles in air is studied. Then, alpha particles are used to measure the thick-ness of an aluminium foil.

A GDM system with a surface-barrier detector; an alpha-radiating source.

Ionising radiation is a way for unstable atomic nuclei to get rid of excess energy. Forsome (usually heavy) nuclei it can be advantageous to emit alpha particles, i.e. 4Henuclei. The resulting daughter nucleus then contains two protons and two neutrons less than the original nucleus. The alpha particles are ejected with specific energies,i.e. only a limited number of alpha energies will occur.

Figure 1 shows a so-called decay scheme, in this case for 235U. The daughter nucleus is usually excited, so that the alpha particles get different energies. The percentage distribution of the occurring alpha energies is shown on the right-hand side.

Figure 1. Example of a decay scheme.

1

1. Investigation of alpha radiation

1.1 Purpose

1.2 Equipment

1.3 Theory

1.3.1 Alpha decay

0.5%

5.7%

0.6%

3.4%

18%

57%4%

1.2%

3.7%

4.6%

92U235

90 Th231

0.448

0.387

0.337

0.279

0.2340.2040.185

0.097

0.042

0

62 0

.201

38 0

.180

50 0

.163

25 0

.110

83 0

.185

17 0

.143

50 0

.204


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The detector used here is of the semiconductor type. Often simply called a semi-conductor detector, the more proper term is surface-barrier detector. It is used to detectcharged particles and measure their energy. These could be alpha or beta particles fromradioactive sources, or other particles, such as protons, produced in nuclear reactions.

The detector consists of an n-doped silicon crystal. By an etching procedure, a very thin(1 µm) p-doped layer has been created at one surface of the crystal. This layer is cov-ered by a thin (40 µg/cm2) gold layer used for electrical contact. When a voltage is applied with the positive potential on the n-layer, the crystal can be regarded as a reverse-biased diode. A depletion layer with very few charge carriers will form in the n-layer. The thickness of this layer will depend on the applied voltage and the electricalresistivity of the crystal. A thin depletion layer is sufficient to detect alpha particles owingto their short range, whereas the detection of beta particles requires a much thicker layer(on the order of mm). The depletion layer of the beta detector used in the present experi-ment is 0.5 mm thick.

The working principle of the surface-barrier detector is shown in figure 2. While acharged particle is being stopped in the depletion layer, it ionises the atoms of the crystal, creating electron-hole pairs. These are collected in the applied electric field,thereby causing a charge pulse.

Figure 2. The working principles of the surface-barrier detector.

If the depletion layer is thick enough, all the kinetic energy of the particle will be deposited, producing a corresponding number of electron-hole pairs. The size of the charge pulse is then proportional to the energy of the particle.

2

1.3.2 The detector

+ -

U

R

Signal

Gold layer

Layer of p-type

Depletion region

Si-crystalof n-type

Charged particle


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The radioactive nuclide used in this experiment is 241Am.

Write down the reaction formula (use the Chart of Nuclides):

.............................................................................................................................................

Use the mass data in the Chart of Nuclides to calculate the energy released in the decay:

.............................................................................................................................................

How many (and which) alpha energies are expected to appear in this reaction, according to the Chart of Nuclides?

.............................................................................................................................................(Compare with the more detailed decay scheme and the table given in the Appendix.)

Place the alpha source in the detector box, as far in as possible. Put the switch on theamplifier in the ”alpha” position. Start the data aquisition. Stop it after a few seconds.Pull out the source 0.5 cm. Start the data aquisition and stop it when the peak in thespectrum is about as big as the first one. Repeat this procedure with the source pulledout in a few more steps of 0.5 cm.

Determine the centroid of each peak and enter the result into the table, together with thedetector-source distance. Remember to add the distance between the source and thedetector with the source pushed as far into the detector box as possible, as an offset tothe measured distances. This offset distance is 1.1 cm for both Gammadata’s alphasource and the Risø source.

3

1.4 Experiment – Stopping power of air for alpha particles

Peak Distance (cm)

Centroid (channels)

1

2

3

4

5

6


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1.4.1 Diagram A – Stopping power of air for alpha particles

Make a diagram of the centroid channel number vs. the distance. Draw a graph that fits the points as well as possible, and extrapolate it to the y axis. The crossing point corresponds to the centroid at zero distance. We can now put an energy scale along the y axis. (The crossing point corresponds to the alpha energy listed in the Chart of Nuclides, and we assume that ”channel 0” corresponds to zero energy. For the Risøsource, which is encapsulated, the energy must be corrected for the energy loss in the coating material.)

Calculate the range in air of these alpha particles by extrapolating the graph down to the x axis.

Answer: ..............................................................................................................................

The slope of the curve gives the energy loss per unit length in air. The energy loss perunit length (dE/dx) is called the stopping power. Calculate the stopping power of air for3.0-MeV alpha particles.

Answer: ..............................................................................................................................

Compare with diagram 1 in the Appendix: ..........................................................................

By calculating the stopping power, i.e. the slope of the graph in diagram 1, at differentdistances from the source, one can construct the so-called Bragg curve, which showsthe stopping power as a function of the distance that the particle has travelled in the substance.

4

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

dE/dx(MeV/cm)

Distance(cm)


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Draw a diagram of the stopping power vs. distance.

As you can see, the stopping power varies with the distance that the alpha particles have travelled. The lower the kinetic energy of the alpha particle, the larger is the stop-ping power. The maximum stopping power occurs at the end of the particle track. Thisappears in the diagram as a peak, the ”Bragg peak”. The Bragg peak means that largeamounts of energy are deposited per unit length towards the end of the particle track.This is used in radiation therapy to damage a tumour as much as possible while sparingthe surrounding tissue.

5

1.4.2 Diagram B – Stopping power of air for alpha particles


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1.5 Experiment – Determination of the thickness of an aluminium foil

Place the alpha source as far inside the detector box as possible. Start the data acquisi-tion and stop after a few seconds. Insert an aluminium foil (mounted in the absorberholder) between the source and the detector. Acquire data again for a few seconds. Determine the centroid channels for the two peaks in the spectrum.

From diagram 2 in the Appendix, the following energy loss per unit length is obtained foralpha particles in aluminium:

dE/dx = .............................. for Eα = ............................. (the average of the energies above)

Calculate the thickness of the foil.(Assume that the thickness is x mg/cm2. Notice the unit! The energy differenceE2 - E1 can be calculated as (dE/dx) • x. Finally calculate the thickness in mm.)

.............................................................................................................................................

.............................................................................................................................................

6

Peak CentroidEnergy

(MeV)

1

2


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To investigate beta radiation with the aid of a semiconductor detector. The spectrum of conversion electron energies is used to determine the energy difference between the atomic K and L shells.

A GDM system with a semiconductor detector; Gammadata’s beta-radiating source.

There is a type of radioactivity in which a nucleus is transformed by emitting beta radia-tion. For instance, 14C decays according to

14C → 14N + β- + ν

This type of instability is due to the fact that the composition of protons and neutrons in the nucleus is not the energetically most favourable one. By radioactive decay the nucleus can reach a state of lower energy. In this process a neutron in the nucleus istransformed to a proton according to

n → p + β- + ν

The total electric charge stays the same, because the positive charge of the proton isbalanced by the creation of a negative particle, β-, which is in fact an electron. An additional particle is also created, a neutrino (in this case an antineutrino, ν).

Another type of beta decay is the transformation of a proton into a neutron:

p → n + β+ + ν

β+ is a ”positive electron”, called a positron. It is the anti-particle of the electron. The released energy becomes the kinetic energy of the created β and ν particles. The energy is arbitrarily distributed between the two particles. The beta energy can get all values from 0 to the maximum energy, which equals the energy released inthe decay. If the daughter nucleus is formed in its ground state, the released energy equals the Q value of the decay.

7

2. Investigation of beta radiation

2.1 Purpose

2.2 Equipment

2.3 Theory

2.3.1 Beta decay

–

–

–


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Figure 3 shows how 137Cs undergoes β- decay to 137Ba. After the decay, the daughternucleus may be in its ground state, but is usually in an excited state with an excitationenergy of 0.662 MeV. This excess energy is emitted either as gamma radiation orthrough so-called internal conversion. In the latter case, the nucleus transfers its excessenergy to one of the electrons of the atom, which is ejected with a kinetic energy of0.662 MeV minus the electron binding energy.

Another example of beta decay (90Sr) is illustrated in the Appendix. 90Sr produces noconversion electrons.

Figure 3. Decay scheme for 137Cs.

The beta particles are detected with the semiconductor detector described in chapter1.3.2 ”Investigation of alpha radiation”.

8

30.0 y

0.662

0

56Ba137

55Cs

137

93.5%

6.5%

2.3.2 The detector

β −


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Place the 137Cs source in the detector box. The switch should be in the ”beta” position. Collect a spectrum for at least 5 minutes.

A beta-ray spectrum is characterised by the continuous distribution of beta-particle energies. The spectrum from 137Cs is dominated by beta particles emitted in decay to the excited state, since 93.5% of the decays go to that state.

The far right end of the spectrum shows a pair of discrete peaks, a large one to the left of a small one. These are due to the conversion electrons that are emitted when the0.662-MeV excited state deexcites. The two peaks show that the conversion electronscome from two different electron shells, the K and L shells in the daughter nuclide 137Ba.

9

2.4 Experiment – Study of beta particles and conversion electrons


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1. Write down the reaction formula for the decay of 137Cs.

.......................................................................................................................................

2. Calculate the energy released in the decay, the so-called Q value.

.......................................................................................................................................

.......................................................................................................................................

3. Calculate the maximum beta-particle energy when the daughter nucleus is left in

a) the ground state: ...........................................................................................................

b) the excited state: ..........................................................................................................

4. Calculate the energy of a conversion electron originating from the K shell. The binding energy of K electrons is 37.4 keV.

.......................................................................................................................................

5. Make an energy calibration of the beta spectrum by means of the ”K peak”, assuming that ”channel 0 = 0 MeV”. Then calculate the energy of the conversionelectrons from the L shell and the binding energy of L electrons.

.......................................................................................................................................

.......................................................................................................................................

6. Calculate the maximum beta-ray energy from the decay to the excited state by ”extrapolating” the right-hand edge of the spectrum distribution down to the energy axis.

Result: ..........................................................................................................................

According to exercise 3b: .............................................................................................

10

3. Exercises

experiments with the alpha/betadetectorlavfis2/bancoapostilasimagens/apradgama... · 2010. 3....

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