EXPERIMENT 21 ALPHA PARTICLES DRAFT AIM

Transcription

1 EXPERIMENT 1 ALPHA PARTICLES DRAFT AIM The aim of this experiment is to investigate the behaviour of alpha particles: their emission, the way they lose energy when passing through media, and the way they scatter from a thin foil. Part of the experiment, an investigation of the scattering of alpha particle from a thin foil, runs overnight and should be set up on your first afternoon in the lab. Leave at least one hour for setting up.

2 PART 1 ALPHA PARTICLE SPECTRA Equipment List:- (a) Vacuum chamber, alpha particle detector, preamp assembly, vacuum gauge and pump (b) Three alpha particle sources (Ra, AMR 13 and AMR 33) (c) Canberra 816 nuclear amplifier (d) Bias supply for alpha detector (e) Multichannel analyzer card (in the PC and run by the Maestro program) 1. THE DETECTOR AND VACUUM SYSTEM The detector is a thin slice of n-type silicon on which is deposited a very thin gold surface, forming a rectifying p-n junction (Knoll 000) The silicon is doped to form a sufficiently wide depletion layer that all the energy of the incident alpha particles is absorbed in the depletion layer, and thus contributes to the resulting voltage pulse. The gold surface also acts as an electrical contact and as a window for the entry of alpha particles into the depletion layer. It is extremely fragile and must on no account be touched. The bias supply for the semiconductor is provided by a 50 volt power supply.. This supply applies a reverse bias voltage to the p-n junction producing a depletion layer about 100 µm thick. The electron-hole pairs formed by ionisation when a charged particle traverses the layer are picked up by the electrodes and input to a preamplifier so that a charged particle results in a pulse being output by the preamplifier. After the preamplifier, there is an amplifier that makes this pulse even larger. The reason we need to increase these voltages, which are inherently small, is to be able to acquire them with data acquisition equipment. The size of the pulses is determined by the gain of the amplifier, which you can control. However, you cannot control the relative sizes of the pulses, which are determined by the relative particle energies. The detector is installed in a vacuum chamber, with the preamplifier mounted above. The preamplifier output is connected to the 816 amplifier, the input of which should be set to receive NEGATIVE UNIPOLAR pulses. To prevent radiation damage to the surface barrier detector, the bias should be switched off and the radioactive source removed immediately counting is completed. The source and detector are placed in an evacuated chamber to avert the interaction of the alpha particles with air prior to their reaching the detector. In order to change radioactive sources (and for part of the experiment on RANGE ENERGY RELATION FOR ALPHA PARTICLES) it is necessary to admit air to the vacuum chamber containing the source and detector, and evacuate the vacuum system after inserting the required source. The following procedures should be adopted to do so:

3 3 EVACUATING THE CHAMBER Close the air inlet valve the valve with the silver coloured knob - close to the mechanical vacuum pump, and turn on the pump, Open the red coloured valve adjacent to the vacuum system to evacuate the chamber. Vacuum valves operate like taps: turn anticlockwise to open, clockwise to close. OPENING THE CHAMBER Air must be admitted into the vacuum chamber before you can open it. Turn off the vacuum pump. Now let air into the chamber by unscrewing the air inlet valve. (Open the red-coloured valve if it is closed.). You should hear air hissing as it rushes into the chamber. Moments later, the top section of the chamber should be easy to lift. When you have completed the experiment, turned off the pump, and open the air inlet valve. This is to prevent oil being sucked back, contaminating the vacuum system and damaging the detector.. CALIBRATION The voltage from the detector is proportional to the energy lost by the alpha particles in the depletion layer. The multi-channel analyzer displays the number of counts corresponding to a particular voltage on the vertical axis versus the amplitude of the voltage pulse on the horizontal axis. Each position on the horizontal axis is known as a channel. We must now obtain a calibration that converts the channel number - which is proportional to the pulse amplitude (strictly voltages between V and ΔV, where the voltage interval ΔV corresponds to the width of a channel ) - into particle energy. Place the AMR33 source (Pu, Am, Cm) on the shelf in the vacuum chamber, close the air inlet valve, start the pump, and evacuate the chamber until the pressure gauge reads 0 Torr. We now need to calibrate the analyzer's horizontal scale in kev per channel. Run the Maestro program to collect the spectrum. The resulting spectrum will consist of 3 peaks corresponding to α-particles of 3 different energies. Note that these peaks are asymmetric: in addition to each strong peak used for the calibration this radioactive source emits a-particles of slightly lower energy than the main α-particle groups. These α- particles take no part in the calibration procedures. The energies of the principal α particles, which are given in Table 1, are used to calibrate the multichannel analyzer.

4 4 Table 1. Energy of principal α particles emitted by AMR33 source Isotope! energy (MeV) 39 Pu Am Cm Set the gain of the 816 amplifier so that the maximum energy of the multichannel analyzer is approximately 8 MeV. (The coarse gain switch should initially be set to 16 and the fine gain set initially to 6.5. It is unlikely that you will need to change the gain during the course of this experiment.) Check that the MeV 44 Cm peak is between channels 1350 and The maxima of the 3 peaks in the AMR33 spectrum are used to calibrate the multichannel analyzer. You can click the mouse on the peak, or move the cursor to the peak by means of the arrow keys, to read the channel number. (At the bottom left-hand corner of the screen.) Take no notice of the in-built calibration in Maestro. You are about to do your own, which will be more reliable. The channel numbers of the peaks in this spectrum and the energies of the corresponding a- particles are used to calibrate the detector/multichannel analyzer system. The response of the surface barrier detector is linear. We only need of these 3 data points in order to produce the required linear calibration of the type E = a + bc p (1) where E is the energy and C p is the channel number of an alpha particle peak. To obtain slightly more accurate calibration data, we will use Origin to fit a straight line through these 3 points. To do so, open Origin (a new worksheet blank table will automatically appear), enter the α-particle energies in Column A and the corresponding peak channel numbers in Column B. Then go to the Tools menu, open Linear Fit and click Fit in the dialogue box that appears on the screen. Values for a and b will be stored in the the Origin Results Log, which will appear on screen when you click Results log in the View menu. (This will mostly appear as a narrow white window immediately below the toolbars, which can be dragged down to view your results.) C1

5 5 3. SECULAR EQUILIBRIUM Remove the AMR33 source and place the Ra source (radium and its decay products) on the shelf in the vacuum chamber. There should be a large peak in the lower channels due to beta particles from the radium and four peaks corresponding to the four alpha particles listed below Save the spectrum onto disk as a user.chn file replacing user with your name and a -3 letters/numerals to identify this file. This should be done immediately on completing the data acquisition. Immediately use the use the Export command from the File menu to convert user.chn file to ASCII format. When exporting a data file, it is not necessary to change the extension to your original file name: a file named user.txt will be created. You will see this file if you view the appropriate directory. Compare the energies of the four peaks with the energies given in (a), (b), (c), and (f) in the table below. Remember, alpha particles will lose energy as they go through matter e.g. gold film, air, etc. When the radium source was prepared it was likely to be almost pure Radium 6. However since that time, some of its daughter nuclides reached secular equilibrium. This is because their half lives are short compared with the time since source preparation. Secular equilibrium is defined for a chain of decays as the situation where all the daughter nuclides have the same activity as the parent. Note that the daughter gas radon produced in Reaction (a), in the table below, is kept in the source by a gold covering. The transitions when radium decays are as follows: Table. Energy of principal α particles emitted by Ra source Decay process Half life (a) (b) (c) (d) (e) (f) (g) (h) (i) 6 Ra Rn + alpha (4.78 MeV) 160 year Rn 18 Po + alpha (5.48MeV) 3.8 day 18 Po 14 Pb + alpha (6.00MeV) 3.05 minute 14 Pb 14 Bi + electron + antineutrino 7 minute 14 Bi 14 Po + electron + antineutrino 17.7 minute 14 Bi 10 Pb + alpha (7.68MeV) 164 s 10 Pb 10 Bi + electron + antineutrino 1 year 10 Bi 10 Po + electron + antineutrino? 10 Po 06 Pb + alpha (5.30MeV) 138 day

6 6 Determine whether the Ra source is in secular equilibrium. Although you can establish this by determining if the peak heights are the same, you cannot do so in practice because three of the peaks overlap and as a result their apparent heights will not be their actual heights. The correct way to do it is to fit Gaussian functions to the peaks and the area under the fitted Gaussians. using the Origin data analysis program.. To load the file into Origin use the Import single ASCII command in the File menu. Two columns should be loaded into the work sheet. The A(x) column is the x-axis (channel numbers) and the B(x) is the number of counts from the detector. Drag the mouse over the A(x) and B(x) in order to select the two columns. Now plot the data by selecting Plot then Line. A plot of the spectrum should appear. Now select the range on the graph that you want to fit by selecting the following button from Origin Arrows will appear on either side of the graph. You can drag these arrows so that they are on either side of the four peaks of the spectrum. Now go to the Analysis menu, then to Fit Multi-peaks, then Gaussian Enter 4 for the number of peaks required. Click OK when the Peak half-width estimate is requested. Now double click the mouse pointer over each peak (the program needs a starting point to its fitting algorithm). A vertical line should appear on each peak. Finally, data should appear indicating the area of each peak, with appropriately fitted lines appearing on the spectrum. If the lines look like they match the spectrum, then you can regard the fit as being good and take the areas as being correct. NOTE: You may use QTiPlot to analyse this spectrum if you prefer. QUESTION: Do you have secular equilibrium? COMMENT. Tutor checkpoint. Obtain tutor's signature before proceeding. 4. RANGE ENERGY RELATION FOR ALPHA PARTICLES Equivalent air thickness of the gold coating of the Ra source The radium source has a 3 µm (nominal) gold-alloy coating deposited over the source to prevent the escape of the radioactive gas radon. Using the range-energy tables for alpha particles in air (given in the Appendix 4), estimate the equivalent air thickness of the thin gold film. This can be calculated for each of the four alpha particles by comparing the range corresponding to the theoretical energy with the range corresponding to the energy observed in the previous section of this experiment. Energy lost by α-particles in air Let air into the chamber until the pressure is atmospheric and take a spectrum for the AMR 13 americium source. If no peaks are seen the source is too far from the counter: the range of the alphas in the air is less than the counter-source distance. If this happens reduce the spacing and repeat the measurements. Evacuate the chamber and measure the spectrum of the AMR13 α-particles in a vacuum and comment on the difference.

7 7 Calculate the effective source-detector separation. Estimate with a ruler the actual separation given the fact that the face of the detector is 89 mm above the bottom sealing flange. Do not attempt to measure the position of the detector with a ruler. Tutor checkpoint. Obtain tutor's signature before proceeding. 5. HIGH RESOLUTION SPECTROSCOPY You will now make high-resolution measurements of the energies and abundances of the alphas emitted from the three isotopes in the AMR33 source used to calibrate the detector. Note that the alphas emitted from these sources are not monoenergetic, but consist of groups of alpha-particles of distinct energy as a consequence of a small fraction of the parent nuclei decaying to an excited state of the daughter nucleus: three for Pu and Am, two for Cm. In order to obtain spectra that enable both the energy and abundance of the alpha particles to be determined accurately it is necessary to obtain the highest counts possible. This means locating the source close to the detector to give a high count rate and counting for as long as possible. Place the AMR33 source (Pu, Am, Cm) in the chamber between 10 and 0 mm from the detector. Evacuate the chamber. Accumulate data until the count in the channels corresponding to the alpha peaks approaches the maximum count that can be recorded by Maestro (,048counts). Calibrate the horizontal scale by taking the energy difference of the main plutonium and curium peaks at and MeV, respectively. Determine the energies and heights of the various α particle peaks by using Origin to fit gaussian profiles to the peaks. Best results are obtained if you analyse the three groups of alphas separately. Compare with the values tabulated values below Compare the full widths at half maximum (FWHM) of the principal peaks in kev obtained from the Gaussian fitting procedure with the resolution located in manufacturers or in texts on nuclear particle detectors. Table 3. Energies and relative abundances of α particles emitted when the parent nuclei in the AMR33 source decay to the ground state, and excited states, of the daughter nuclei Parent nucleus Daughter nucleus Energies and abundances of the main groups of particles 39 Pu 35 U (7%) (17%) (11%) 41 Am 37 Np 5.48 (85%) (13%) (1.7%) 44 Cm 40 Pu (75%) (5%) Tutor checkpoint. Obtain tutor's signature before proceeding.

8 8 SUPPLEMENTARY QUESTION (These questions are quite advanced and you may like to consider them in a report if you have time to do the required research). Various factors determine the probability of α decay. If you look at the reference book Table of Isotopes you will be able to observe the following general tendencies. a) The higher the energy of the emitted α, the higher the probability of that transition. b) The larger the spin change of the parent to daughter nucleus the lower the probability of the transition. c) Transitions with no spin change must be between states of the same parity. Can you see the theoretical reason for the above? Find out about the "Coulomb Barrier" and the "Centrifugal Barrier" as applied to alpha decay. Remember the spin of the alpha is zero and its parity is positive. For further help consult the following references: Reference for Geiger Nuttall law: RB Leighton, Principles of Modern Physics. Reference for Centrifugal Barrier: AES Green, KS Krane, Introductory Nuclear Physics See also WE Burcham, Elements of Nuclear Physics

9 9 PART ALPHA PARTICLE SCATTERING Equipment List:- (a) (b) (c) (d) (e) (f) (g) Scattering chamber: thin gold foil, and alpha detector in an evacuated chamber. Preamplifier located at the scattering chamber NIM bin with Canberra 1431 Single Channel Analyser Detector bias supply Surface Barrier Detector Computer Interconnection cables This equipment is located on a bench behind the apparatus for Part 1 of this experiment 1. INTRODUCTION Around 1910, Rutherford and co-workers showed that the nucleus of an atom was extremely small around 10-4 of the size of an atom by observation of the scattering of alpha particles off a thin gold foil. In this experiment Rutherford s work will be reproduced using modern equipment.. THEORY Let us assume that the nucleus is a small sphere of radius R with a charge + Ze and we bombard a thin foil with alpha particles of mass m, charge + Z! e, travelling with initial velocity v. ( Z! = for alpha particles). Some of these alpha particles will be scattered as they pass through the foil. We will assume that: (a) v = c so that a non-relativistic treatment can be used. For alphas emitted from radioactive elements, v/c is less than so this is reasonable. (b) (c) (d) The nucleus is fixed (instead of recoiling). Rutherford et al used a gold foil where the atoms were 50 times heavier than the alpha particle; so this assumption is reasonable. The apparatus here also uses a gold foil. The alpha particle never scatters more than once when passing through the foil. You will see that this is justified when you do the experiment: even for small angles of scattering, the probability of scattering is less than so that the probability of an alpha particle of scattering twice is less than one part in a million. The alpha particles do not scatter off electrons. This is not strictly true but the angles of scattering are small because the ratio of the electron mass to that of the alpha is about We consider the nucleus to be a small sphere. The force (Coulomb force) between it and the alpha is repulsive and follows the inverse square law, causing the path of a scattered alpha to be a hyperbola. F ( Ze) ( Z# e) = () 4! " r 0

10 10 The angle between the asymptotes of the hyperbola is the angle of scattering θ. This latter depends on a quantity called the impact parameter p. This is the perpendicular distance from the nucleus to the line that the alpha would have followed if it had not been scattered. (See Figure 1 below.) Asymptote!'s path P Gold Nucleus F " # 0 " A G (focus) p Asymptote Figure 1. Geometry for the scattering of an alpha particle off a nucleus. Because the Coulomb force between the two particles is central, i.e. along the line joining them, we can say that the angular momentum of the alpha is conserved in its orbit around the nucleus. With this and the conservation of energy, we show that p and θ are related by p ( Ze)( Z$ e) cot (! / ) = (3) 4"# mv 0 This relationship between impact parameter and the scattering angle is derived in Appendix 1. It is of course impossible to aim alphas at a nucleus to get a given value of p - alphas can go anywhere - but we can mentally draw a ring of radius p and width dp around each nucleus and handle the problem statistically. The result is that the rate of detection of alphas is given by (! ) A R0 Nt 4 R( " ) = B cosec (" / ) (4) 4r where R 0 is the number of particles bombarding the foil per unit time, is the area of the detector at a distance r from the foil, N is the number of atoms per unit volume of the foil and t is the foil thickness. This expression - the Rutherford scattering law - is derived in Appendix. Thus R (! ) is proportional to: 4 (a) cosec (! / ) (b) R 0 (c) t (d) 1 ( mv ), i.e. to the inverse square of the alpha particle s energy. The importance of (c) and (d) will become evident later on. For this part of the experiment equation (3) may be rewritten as

11 11 ( ) 4 R (! ) = cosec! / (5) 3. OBSERVATIONS These are done with an apparatus described by PJ Ouseph (1976). (You might also like to read Ramage et al. (1975).) The scattering is done in a vacuum chamber containing a collimated source of alphas (Americium 41) aimed at a 0.5 µm ( g/cm ) gold foil. A surface barrier semiconductor detector registers the scattered alphas at an angle θ which is varied by a stepping motor under computer control. The foil is turned with gears so its thickness t decreases with θ as = t 0 sec ( ) so that the effective thickness of the foil is independent of the scattering angle. Evacuate the chamber. Do not change the settings of the SCA 1431 single channel analyzer. The experiment is controlled by a dedicated computer. You will be using the program Alpha Particle Scattering to take measurements over as long a period as possible (usually over night). The program contains various sections, each identified by a "button" on the display screen. These are: o Hardware Test. This section of the program checks the operational status of the counters, the stepping motor and the steering mechanism. o Manual. This section enables you to control the position of the detector and the counting time manually. o Run. In this section the computer controls both the position of the detector and the counting time to: Step the counter through the zero position, observing the counts for each angle, which enables the computer to estimate the true zero position (for undeflected alphas) of the detector Record the count rate as a function of the angular position of the detector. o Enter and plot corrected data. As Origin or QTiPlot are now used to record and analyse the data, this section is no longer used. Click on the Hardware Test button to enter this section, and check the operation of the detector and the stepping motor that controls the position of the detector. Using Manual control of the system, set the detector to 10 and count for 10 minutes. From the measured count rate determine the constant C in equation (4) Use equation (4) to determine the counting time required to obtain an accuracy of 5% in the count rate when the detector is set at 40. Recall that in a counting experiment Accuracy = Standard deviation of count N 1 = = Count N N where N = Total count recorded Z! = It should be obvious from this calculation that an extremely long counting time would be required for high accuracy at large scattering angles.

12 1 Now click on the Run button and then the Calibrate button. This steps the counter through the zero position observing the counts for each angle, which enables the computer to more accurately estimate the true zero position (for undeflected alphas). All subsequent settings must be corrected for this zero offset. Print out the results of this calibration procedure. After determining the zero offset click the Continue button to proceed to the computercontrolled measurement of the count rate as a function of scattering angle and a blank table will appear on screen. Enter the counting times for various scattering angles given in Table 4. Verify that you have entered the correct values then start these measurements. Table 4. Recommended counting times for specific scattering angles Scattering angle (θ) (degrees) Counting time (minutes) Place a notice adjacent to the experiment to advise that an overnight run is in progress and notify Marek Dolleiser (the Laboratory Senior Technical Officer) that you have started an overnight run. (Marek will usually download your results the following morning.) After your run, obtain a print-out of your data and use Origin to plot the final data in log-log format to determine the slope. From the slope, determine whether your observations of count rate as a function of scattering angle are consistent with equation (4)above equations are followed. If you have time, you may wish to correct your raw data for: o The finite width of the (imperfectly) collimated alpha beam and the finite solid angle subtended by the detector o The finite width of the (imperfectly) collimated alpha beam and the finite solid angle subtended by the detector at the gold foil The procedures for adjusting your raw data to take account of these errors are discussed in Appendix 3. All the computer instructions are described in greater detail in accompanying bench file. Tutor checkpoint.

13 13 References Knoll Glen F (000) Radiation Detection and Measurement, (New York: John Wiley) Ouseph PJ (1976) Am J Phys 44, 101 Ramage JC et al (1975), Am. J. Phys. 43, 51

14 Appendix 1 14 Derivation of the relation between impact parameter p and the scattering angle θ. Because the force between the two particles is inverse square repulsive we can say that the path of the alpha will be a hyperbola with the gold nucleus at one focus. Asymptote P!'s path Gold Nucleus F " # 0 " A G (focus) p Asymptote. Figure 1. Geometry for the scattering of an alpha particle off a nucleus. (Repeated for your convenience.) Figure scattering of n a particle off a thin gold foil. In the figure we have: o! = 180 # " FP - PG = k (a constant) } FA - AG = k (the same constant) a property of a hyperbola Similarly cos! = k a where a = FG. (We have just sent P off to infinity along an asymptote!) Finally we have p = FO sin! = a / sin". By elimination we obtain FA (which we'll call r), which is given by ( tan( ) sec( ) ) r = p! +! (6) If the velocity of an α particle at is v and the velocity at A (where the a particle is closest to the nucleus) is w, then the conservation of angular momentum gives mvp = mwr (7)

15 From conservation of energy 15 (! " ) = + # 0 (8) 1 mv 1 mw Z Z e 4 r Equations (7) and (8), together with equation (6), which links r and p, provides the required result by elimination of w and r. Z Z$ e cot(! ) p = (9) 4" # 0mv Appendix Derivation of the Rutherford scattering law The analysis presented in this Appendix is applicable to any situation where one particle (molecular, atomic or nuclear) is scattered from another. If we mentally drawn a ring of radius p and width dp around each nucleus in the scattering foil, the area of this ring is the differential cross-section for the scattering, d!. Differentiating p we obtain dp Z Z$ e cosec (! ) = d! (10) 4" # 0mv The negative sign can be dropped because cross-sections cannot be < 0. Such an encounter will result in the alpha being scattered through an angle between θ and dθ. The scattering foil can be pictured as having N points per unit area representing the nuclei in it. Around each one there is a "ring" of radius p and width dp. Thus the probability of a scattering through an angle θ to θ + dθ can be written ( ) dp = N# d! = Nt d! = Nt " p dp where N is the number of nuclei (= number of atoms) per unit volume in the foil and t is the foil thickness. To detect the scattered alphas we can use a detector of sensitive area δa at a distance r such that! A = 4" r. Considering δa to be the small annulus lying between θ and θ + dθ we find the following relation between δa and dθ A = r sin d! " # # = ( ) ( ) 4" r sin # cos # d# The rate of detection of alphas will be R(θ) = R dp where R is the number of alphas bombarding the foil per unit time and, from equation (10), R #! A R Nt $ = % & ' 4r ( (" 0 ) B F (" ).

16 16 (! ) (! ) (! ) (! ) cot cosec where F (! ) = = cosec! sin cos and B = Z Z$ e 4"# mv 4 ( ) B has the dimensions of length. It is in fact half the distance of approach of the alpha particle to the nucleus when θ = 0 ie the distance of closest approach. For 10 MeV alphas on gold, this distance is greater than the nuclear radius so that the alphas do not see the nuclear force. If this was not so then there would be a deviation from the Rutherford formula at large θ. a) Correction to Angles Appendix 3 Corrections to Data The finite width of the (imperfectly) collimated alpha beam and the finite solid angle subtended by the detector mean that the scattering angle as set is not the same as the "effective" scattering angle. If we assume that the distribution function describing the incident beam and detector widths is triangular in form, Tri(θ), then the effective angle is then the mean of the distribution Tri(θ)cosec 4 (θ/). We have this computed using Simpson's rule. The result, in the form of a correction to the set angle, is given in the following table. Table 5. Correction (in degrees) to be subtracted from set angle to obtain effective angle. Set Angle Half Width of Distribution 4 deg 5 deg 6 deg 7 deg 8 deg 9 deg Corrections to be applied The computer s printout of the graph drawn when θ = 0 was found may be used to correct for the imperfect collimation of the alpha particle beam and the finite acceptance angle of the detector. Fit visually i.e. by trial and error - a triangular distribution to this graph. The half width of this distribution, which is required to apply the correction discussed above is half the base of this triangle.

17 17 b) Correction to Rates A correction has also to be made to the rates. Since the scattering foil has a finite thickness the alphas have less energy at the point of scattering than they have when they enter the foil. This is because of the interaction of the alphas with electrons. Ordinarily this would cause no complication but as mentioned above the foil is turned by gearing so that its thickness t increases with θ. The average energy of scattering thus decreases with t. Recall that the rate of detection of alphas is proportional to the inverse of the this energy. In addition we have, in effect, a thicker foil at larger angles (which will conversely increase the rate). Both these affects can be allowed for with a single correction term applied to the rates of "! # ft cos 0 % & $ ' ( E where E 0 is the incident energy of the alpha and f is a constant which we now define. 0 The energy loss of the alphas can be approximated by the formula constant for each absorber and incident particle.! de f dx = E where f is a This integrates to 0 E E ft =! and using the dependence t t (! ) = 0 sec this becomes ( ) = 0 " 0 sec for scattering occurs on the far side of the foil and E E ft! ( ) = " when the scattering occurs halfway through the foil, which we will E E 0 ft 0 sec! use as an average scattering event. Since, from the Rutherford formula R! t E we have R obs (! ) sec(! ) cos(! ) t0 sec t0 t " = = 0 E0 % ft0 E0 % ft # 0 ft $ E 0 0 & cos(! ) % ' &( E 0 ') Thus Robs = Rtrue "% cos(! ) $ 0.045# & The value of f is approximately 1500 MeV cm g -1 ft and thus the factor 0 E is approximately = obs "% cos $ 0.045# & so that R R (! ) true This means one should obtain a line a slope -4 on a log log plot of R true against ( ) sin!. where θ is the corrected scattering angle as discussed earlier in this Appendix.

18 18 Appendix 4 Table 6. Alpha particle energy vs. mean range in air at 15ºC and 760 mm Hg Mean range (cm) Alpha energy (MeV)