In this lesson you will use the Cloud Chamber applet to investigate the alpha decay process.
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1 Alha Decay In this lesson you will use the Cloud Chamber alet to investigate the alha decay rocess. Prerequisites: You should be familiar with the hysics of charged articles moving in magnetic fields and know how to use the hand-rules. You should know what an alha article is Learning Outcomes: When finished you should be familiar with the alha decay rocess including the tyical energies emitted during alha decay. Oen the Cloud Chamber alet in the toolbox on the right and use it as you answer the following questions. In this alet, the magnetic field is assumed to oint out of the screen. The strength of the magnetic field is shown on the info anel on the right hand side of the alet. Be sure to select "Alha decay" mode (under the "Otions menu", select "decay" and then click on "Alha"). Investigating the Decay of Americium Use the alet to determine whether alha articles are ositively or negatively charged. xlain in a well reasoned sentence how you determined this. 1
2 . Adjust the magnetic field strength until you get strongly curved article aths. Do this by selecting the "Advanced Settings" choice from the Otions menu. What are "tyical" magnetic field strengths that you will need to roduce measurable radii for the alha article aths? 3. True or False: An alha article is a helium ion. xlain your answer. 4. In a revious unit you learned that a magnetic field of strength B exerts a force on a moving article of charge q. If the article is moving at right angles to the field then the magnetic force is given by the exression Fmagnetic = qvb. xlain why this creates a circular arc and use your knowledge of circular motion to show that the momentum of the article can be written as = Bqr, where is the magnitude of the momentum and r is the radius of the ath's arc.
3 5. Show that the kinetic energy of a article can be related to the momentum by the equation k = and that the kinetic energy of a article can be m related to the radius of its arc by the exression k B qr =. m 6. Produce 5 different aths for the alha decay of Americium-43 and record your results in the table below. Measure the radius of each ath as carefully as you can and use this to hel determine the kinetic energy of alha articles emitted by Americium-43. The mass of an alha article is x 10-7 kg and the charge is +e. In the last column of the table you are asked to convert this to units of MeV or Million lectron Volts. Use the conversion 1 MeV = 1.60 x J. Trial # Magnetic Field (T) Radius (m) nergy (J) nergy (MeV) Average energy in MeV: 3
4 Summary Americium-43 (Am-43) is a very commonly used radioisotoe you robably have some in your home! This isotoe is used in smoke detectors. Your measurements should have shown that the energy of alha articles emitted during the alha decay of Am-43 is aroximately 5.4 MeV. Am-43 has a halflife of about 7370 years. Comaring Alha Decay of Various Nuclei We have examined alha decay in one articular nucleus, Americium-43. Do alha decays behave the same way in other nuclei as they do in Am-43? In this section, we will discover the answer to that question. 1. Use the alet to comare the energies of alha articles emitted by the nuclei listed below (to change the tye of nucleus, go to "Advanced Settings" under the Otions menu). Run at least 3 different trials for each nucleus and average the results. Choose a convenient magnetic field strength to allow you to make an accurate measurement of the ath radius. Using the same method as in the exloration of Americium-34, exress the energies in MeV. Nucleus # of Trials Average Radius (m) Average nergy (MeV) Americium-43 Thorium-3 Uranium-35 Radium-6 Polonium-10 4
5 . Was there much variation in the energy of the alha articles that you measured for any one element? To what would you attribute any variations in energy that you may have detected? Summary Alha decay is one of the rimary modes of radioactive decay. For very large nuclei with many neutrons and rotons, it is ossible for rotons and neutrons to bind together, inside the nucleus, as an alha article. While inside the nucleus, the alha article is bound to the nucleus through the Strong Nuclear Force. The debroglie relation, however, reminds us that articles, including alha articles, have wave roerties. One of these wave roerties leads to a rocess known as quantum tunnelling, in which there is, deending on the binding energy of the nucleus, a robability of the alha article existing outside the nucleus that is, it can escae from the nucleus and aear to be emitted. When this haens, a small amount of the mass of the original nucleus is converted into energy. You can calculate the amount, designated as Q, by noting that the sum of the masses of the final (or daughter) nucleus and the alha article is a little less than the original (or arent) nucleus. This can be exressed via the equation Q = ( M M M ) c. D When an alha article is emitted, the laws of conservation of energy and momentum must be obeyed. This means that the alha article and the arent nucleus must have equal and oosite momenta. The arent nucleus recoils when the alha article is emitted. Since the arent nucleus and alha article have equal and oosite momenta, we can use this to determine how much kinetic energy will be carried away by the alha article. Recall that momentum 5
6 and kinetic energy are related via the exression k =. Since the energy that m is released ("Q") includes the kinetic energy of both the recoiling daughter nucleus and the alha article, it can be written as Q = M +. By dividing M everything by, the relation becomes Q = 1+ M M M D D M. Since alha decay reduces the atomic mass number ("A") by 4 units, this equation can be re-written as k Q( A 4) = =. M A You can use the above to redict the kinetic energy of an alha article emitted during the alha-decay of Uranium-38 into Thorium-34. To see how, start with: U Th Use Q= M M M ) c ( D and insert the aroriate numbers to get: MeV Q= = c (Note that MeV/c units are used here for the masses). ( ) c 4.3 MeV This means that the emitted alha-article will have a kinetic energy of k ( ) QA ( 4) 4.3 MeV 38 4 = = = 4. MeV. A 38 6
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