EXPERIMENT 13. Radiation Laboratory Neutron Irradiation. Introduction. Background. Neutron Activation Equations

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1 EXPERIMENT 13 Radiation Laboratory Neutron Irradiation Please Read the Radiation Laboratory Safety Regulations at the back of this book The purpose of this experiment is to examine neutron slowing down in paraffin wax and determine the slowing down length by measuring the neutron radial flux distribution in a fairly simple geometrical set up. Paraffin wax is a saturated aliphatic hydrocarbon of the methane series having the general formula C n H 2n+2 with the value of n being typically about 25 (eg. C 25 H 52 ). Introduction Neutron activation analysis is a very powerful technique for identifying many elements. Basically the technique is quite simple. A sample is irradiated by slow neutrons and becomes radioactive. By measuring the radioactivity (β + s, γ s, β s) and half-life of the resulting sample the elemental constituents of the sample and their relative concentrations can be identified. Industrial activation analysis is usually done with slow neutrons from a reactor, where the neutron flux can be as high as neutrons/cm 2 /s, or with an accelerator with fast neutron fluxes of neutrons/cm 2 /s. When activation analysis is compared with other instrumental analyses such as gravimetric, colorimetric, spectrographic, or mass spectroscopy, its sensitivity is usually shown to be better by a factor of 10 than that of other methods. Activation analysis is used extensively in such fields as geology, medicine, agriculture, electronics, metallurgy, criminology, and the petroleum industry. Background Neutron Activation Equations Assume that the sample has been activated by a neutron source. At the instant when the activation has been terminated (t = 0), the activity of the sample is given by the following expression: 13-1

2 where A 0 = σmηφαs w (13.1) A 0 = the number of disintegrations per second of the element in the sample at t = 0 (when irradiation stops), σ = cross section for the reaction, cm 2, (: Note that the term cross-section in nuclear physics refers to the probability of an interaction occurring when a particle hits a target) m = mass of the target element, grams, φ = neutron flux, neutrons/cm 2 /s, η = Avogadro s number, molecules/mole, α = fraction of the target isotopes in the sample [e.g. with the Vanadium sample producing the V 51 (n, γ)v 52 reaction, α =.997 since 99.7% of all natural Vanadium is V 51 ], S = saturation factor, 1 - e λt, where λ = 0.693/T 1/2 and T 1/2 is the half life for the reaction, w = atomic weight of the element. (Note : In the above definitions, t and T 1/2 must be in the same time units.) The values to be used in this experiment for the reaction V 51 (n, γ)v 52 are: σ = 5.0 x cm 2 m= 8.5 grams α = S = 0.9 and w= Let us examine Eq in terms of our knowledge about a reaction. For example, if we were activating the Vanadium sample, the following reaction would take place: V 51 (n, γ)v 52, T 1/2 = 3.76 minutes The cross section is 5.0 x cm 2. For our example then, we can determine everything in Eq except A 0 and φ. A 0 will be measured with a scintillation counter and you will determine φ in the last part of this experiment. Table 13.1 is a list of common thermal neutron cross sections. 13-2

3 Table 13.1: Common Thermal Neutron Cross Sections Al(n,γ) 28 Al ± barns V(n,γ) 52 V 5.00 ± Cu(n,γ) 64 Cu 4.51 ± Na(n,γ) 24 Na ± Mn(n,γ) 56 Mn 13.3 ± 0.2 Note : 1 barn = cm 2 Neutron Source and Neutron Activation A 1 Curie or 37 Giga Bequerel (37 GBq) Plutonium 239 source surrounded by a sheath of Beryllium is situated at the centre of a tank of paraffin wax (see Fig. 13.1). The Pu 239 nuclei decay with a half-life of years ( seconds) and a release of 5.6 MeV mainly in the form of alpha particles according to the reaction Pu U α (5.5MeV) The alpha-particles released interact in the Beryllium target to produce fast neutrons of energy 5.7 MeV. Be He 4 2 C n(5.7mev) The total neutron flux produced by the source (Figs and 13.2) used in this experiment is approximately fast neutrons per second. Certain materials when placed in a neutron flux undergo neutron activation reactions i.e. a neutron is absorbed and a gamma-ray emitted. In this experiment the material used is Vanadium and the activation reaction is: V 51 + n V 52 Cr 51 + β + γ The V 52 decay with a half-life of about 3.76 minutes to a stable nuclide with the emission of a beta particle and gamma radiation of energy 1.43 MeV. Measurement of the gamma radiation emitted from a sample of Vanadium gives the number of V 52 states produces by irradiation in the neutron flux. Due to the half-life of the gamma emission the number of V 52 nuclides reaches a saturation value after about 10 minutes. Therefore the number of gamma-rays emitted from the Vanadium sample after 10 minutes irradiation is a direct measure of the slow neutron flux at the position of the sample irradiation. Detection of Gamma Radiation from V 52 The detection system for the gamma radiation is identical to the experiment Gamma-Ray Absorption and Counting Statistics. You should review the instructions for this experiment and the 13-3

4 Handle Paraffin wax Source in raised position 97 cm Paraffin wax Vanadium sample holder 110 cm Figure 13.1: The large paraffin wax tank containing the Plutonium source. By turning the handle the source is raised to the level of the sample holder. instructions for the operation of the computer programs. The thallium-activated Sodium Iodide detector and the photomultiplier are mounted in a lead shield to reduce the background gamma radiation from the walls and the cosmic radiation. Experimental Procedure You need to locate the photopeak for the 1.43 MeV gamma radiation. Place the sample in the holder (see Figs and 13.2) at the closest position (r = 3 cm) to the neutron source. Irradiate the sample for about 5 minutes. Remove the sample and place it in the sample holder located below the sodium iodide detector. Run the PHA2 program and locate the photopeak at about channel 850. Place the cursors at around channels 800 (839 New App) and 900 (951 New App). You are now ready to proceed with the experiment and count the monoenergetic gamma-rays in the photopeak. To save time please note that there are at least 6 Vanadium samples all with equal mass, that can be loaded into the paraffin wax tank and irradiated simultaneously. All samples should be irradiated for at least 10 minutes. The slots in the sample holder are at r = 3, 5, 7, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22 and 24 cm from the neutron source. (a) Record the count (N) in the photopeak every 30 seconds for four minutes for the irradiated sample from r = 3 cm. This sample should have been irradiated for 10 minutes. Record the background count from a non-irradiated sample over the same period. Over the 4 minute period, the count in the photopeak will be about 400 and the background will be about 25. When you subtract the background you must add the errors in quadrature i.e. (400 ± 20) (25 ± 5) = 375 ± = 375 ±

5 Top View Figure 13.2: A top view of the parafin wax tank showing the location of the sample holder for the Vanadium samples. (Just as in the Brownian Motion experiment, the error on the counts is simply the square root of the counts. Check your notes from the introductory laboratory lecture). Plot the background-subtracted counts as a function of time from irradiation on semilog paper and determine T 1/2 from the best fit slope of the line. The value of T 1/2 is 3.76 minutes. Do this in Mathcad or Matlab. Recall from 1st year that N N 0 = e λ t where N = counts N 0 = counts at t = 0 λ = decay constant Therefore, by plotting ln N versus t, the slope yields λ. T 1/2 = ln 2 λ hence the half-life T 1/2 can be determined. From the error on the best fir slope, also determine the error on T 1/2. (b) 1. The Vanadium samples are irradiated at the known distance for at least 10 minutes. 2. Each sample is removed from the neutron tank, dried and placed in the holder under the Sodium Iodide detector. 3. Thirty seconds after removal of the Vanadium sample from the neutron tank the count in the pulse height analyser is recorded every minute for 4 minutes. Let N be the count in the four minute period. The background count obtained in part (a) is subtracted from N. 13-5

6 Figure 13.3: Graph of r 2 N(r) versus r. 4. The irradiation and counting is carried out for the radial positions between r = 3 cm and r = 24 cm. At the larger distances (r 20 cm) the count will be approximately equal to the background. Determination of neutron slowing down length A graph of r 2 N(r) versus r is plotted. It should be of the form shown in Fig The maximum of the curve occurs at a radial distance r 12 cm. The shape of the distribution in Fig arises from a number of complicated processes. Beyond the maximum the distribution is experimentally decreasing as e r Ls where Ls is the Neutron Slowing Down Length in paraffin wax. A semilog plot of part (2) of the distribution is made. The slope of this graph is 1 and hence Ls Ls can be determined. Make sure an adequate number of experimental points are obtained for a good plot of part (2) of the distribution. Your result for Ls should be between 10 and 20 cm. Estimate the error on the slope determination and hence on Ls. Determination of Scattering Cross Section The scattering cross section δ SC of the neutrons by the protons in the paraffin wax is the effective area of a neutron if we regard the proton as a particle of zero area. If the paraffin wax contains n protons per cc. the average volume per proton is n 1 cc. If the average distance travelled by a neutron before an interaction is Ls then the volume swept out is Lsδ SC. Since the volume on average contains one proton then nlsδ SC = 1. Each molecule of paraffin wax contains about 52 H atoms (protons). The chemical formula of the wax is C 25 H 52. Calculate the number of protons from H per unit mass. The density of the wax is approximately kg/m 3. Hence calculate n, the number of protons from H per cc. Hence 13-6

7 δ SC can be calculated. Due to the relative masses of the neutron and the carbon nucleus the carbon is ineffective in slowing down neutrons and only the contribution from H atoms is considered. Determination of the thermal neutron flux (φ) at the radial position r = 12 cm The data required for the determination of φ was obtained previously. After irradiation the sample was transferred immediately to the Nal(Tl) detector, and a spectrum accumulated for a time (t 1 ) long enough to get reasonable statistics under the photopeak. The time is usually at least one halflife. The true number of disintegrations (N d ) that occurred during t 1 can be determined from the following ρ β N d = (13.2) Gε ρ f where ρ = sum under the photopeak for r = 12 cm, β = background for the same counting period under the photopeak, G = A/4πs 2, where A = area of detector, in cm 2, and s = distance from source to detector, ε ρ in cm. (Note A = 20 cm 2, s = 2.5 cm, and hence G = 0.25) = intrinsic peak efficiency for the gamma energy and the detector size used. (Use a value of 0.01 for ε ρ ), If using new apparatus this value should be used f = decay fraction of the total disintegrations in which the measured gamma ray is emitted. Use f = 1.0. From the decay equation, N 0 can be calculated : N d = N 0 (1 e λt ), A 0 = λn 0, where t is the time for which the sample was counted. At r = 12 cm, your value of A 0 should be about 500 disintegrations per second. Therefore we have reduced Eq to one unknown, φ, which is the number of neutrons/cm 2 /s for the neutron source. Your value of φ should be about 1000 neutrons/cm 2 /s at the radial distance r = 12 cm. Appendix Neutrons are usually divided into three classes depending on their energy E: Slow neutrons E < 1000 ev Intermediate energy neutrons 1 kev < E < 500 kev Fast neutrons 0.5 MeV < E < 20 MeV 13-7

8 Fig shows the total cross sections σ t of hydrogen as a function of neutron energy E. Note that the cross sections is given in barns where 1 barn = cm 2. The cross section is constant between 1 ev and 1 kev but decreases rapidly with energy above E = 0.01 MeV. A large number of studies have been performed on thermal neutrons at large distance from a point source in water. (Note that paraffin wax is a slightly better absorber than water. Why?) Typical results are shown in Fig where the quantity r 2 N(r) has been plotted on semilog paper. All curves show a linear decrease at large distances r from the source (r > 12 cm). The explanation of this behaviour is straightforward if we note that the neutron scattering cross- section of hydrogen falls off very sharply with energy (Fig. 13.4). If a neutron emitted from the source once collides and experiences a large energy loss, the cross-section for additional collisions is large. Thus the probability of the neutron being moderated and finally absorbed near the site of its first collision is large. The thermal and epithermal fluxes of neutrons that are detected by the Vanadium should behave like the flux of primary neutrons. Figure 13.4: Total Cross Section of Hydrogen as a function of neutron energy. Question 1. Why is the carbon nucleus ineffective in slowing down neutrons? WWW : Institut Laue-Langevin neutron facility