CLASS 3. NUCLEAR BINDING ENERGY 3.. INTRODUCTION Scientists found that hitting atoms with alpha particles could induce transformations in light elements. (Recall that the capture of an alpha particle by a nucleus changed the identity of the target atom). These experiments were limited because the alpha particles were obtained from the natural radioactive decay of elements. There was little control over the speed of the particles or the number that could be obtained, and only a few other particles besides alpha particles could be used. Alpha particles are light, the available beams had low intensity (i.e. few particles) and each alpha particle moved slowly. Alpha particles aren t powerful enough to induce transmutation in heavier atoms. The ability to investigate heavier nuclei (which require more energy to break apart) necessitated the development of machines that could provide scientists with a larger range of projectile particles. Much of this new work was being done in the United States. After World War I, administrators realized the importance of science and technology for the national defense (as well as the economy), and started putting money into scientific research. 3.. GOALS Explain why how particle accelerators work and why they were needed to continue our learning about the nucleus Be able to convert between electron volts and joules Understand that stable nuclei stay together because the energy they have as one nucleus is less than the energy they would need to exist independently. Explain what E=mc means. Explain what binding energy per nucleon is and use a charge of binding energy per nucleon vs. mass number to determine whether an element is likely to be stable Use a chart of number of neutrons vs. number of protons to determine whether an element is likely to be stable and, if not, what type of decay it is likely to undergo. 3.3. PARTICLE ACCELERATORS 3.3.. Units. It is useful here to talk a little about a new type of unit for measuring energy. The electron-volt (ev) is the energy it takes to accelerate an electron through a potential difference of volt. It is a much smaller unit than the joule, which is appropriate when dealing with atomic and nuclear physics. The conversion factor is: -9 ev =.60 0 3.3.. The Need for Particle Accelerators. What are the differences between heavier atoms and lighter atoms in terms of using them as projectiles? One difference is that the nucleus of a heavier atom has a larger positive charge. Coulomb s Law is: qq F = k r where q is due to the alpha particle and q to the nucleus of the atom that is going to be hit. When q increases, the force F between the two particles increases and the alpha particle is repelled. In lighter elements, the α particle is going fast enough to overcome the electrical repulsion. In heavier elements, the force is larger (because the nucleus has more charge) and the α particle cannot overcome the repulsion of the nucleus. Particles like the proton or deuteron have a small positive charge and are less massive. This makes them easier to move, and the electrostatic repulsion (as described by Coulomb s law) is reduced because the charge is smaller. This increases the likelihood of a reaction between the particle and J
the nucleus. It would be helpful for the scientist to be able to control the energy (speed) of the particle but there was no way of doing this with the radioactive materials known at the time. 3.3.3. Linear Accelerators. Charged particles can be moved by electric and magnetic fields. In 930, John D. Cockcroft and Ernest Walton at the Cavendish Laboratory (Cambridge, England) built an accelerator that used a very high electrical potential to accelerate charged particles down an evacuated tube. The tube was straight and the material to be studied was placed at the end of the tube. Their design allowed them to accelerate particles to very high speeds. The difficulty was that it required a very large electrical potential. It is difficult to produce a large potential difference, so the maximum energies of the particles were limited. In 93, Crockoft and Walton succeeded in producing the first transmutation involving protons, which was converting Li-7 into two alpha particles: 7 4 4 3Li + p He + He. The tube they accelerated the protons down was about 8 feet long. 3.3.4. Cyclotron Accelerators. Ernest O. Lawrence (University of California - Berkeley) realized that linear accelerators might be a possible solution for accelerating heavy atoms; however, if you wanted a beam of fast-moving alpha particles, the length of the accelerator would have to be many, many meters long. Lawrence, along with chemist Gilbert Lewis and electrical engineer Leonard Fuller, invented the cyclotron accelerator (or just cyclotron ). As in the linear accelerator, groups of particles are pushed by electric fields and accelerate. In the cyclotron, magnets ( Dees ) are used to bend the particles and move them around a ring instead of through a linear tube. Each time the particles circle the ring they get another push, so you can choose what energy you would like the particles to have. Figure 3. shows a schematic for the Stanford Linear Accelerator. The ions that are being accelerated start in the center and spiral outward. A graduate student, M. Stanley Livingston, did much of the work of translating Lawrence s ideas into an actual piece of working apparatus. In January Dr Livingston has asked me to advise 93, Lawrence and Livingston produced the first you that he has obtained,00,000 cyclotron, a device about 4.5 inches in diameter that used volt protons. He also suggested that I repeated passes through a potential of,800 volts to add Whoopee! Telegram to accelerate hydrogen ions up to energies of 80,000 electron Lawrence, August 3 rd, 93 + Power Supply Figure 3.: A linear accelerator uses a potential difference to accelerate a charged ion. Figure 3.: A cyclotron accelerator (Diagram from the Stanford Linear Accelerator). The electric fields used are radio-frequency electric fields, not dc electric field
volts. In summer 93 an eleven-inch cyclotron achieved a million volts. Since the energies of the particles increased as larger cyclotrons were built, Lawrence proceeded to make a 7 and then a 37 version. A 84 cyclotron was completed in 946. Lawrence received the Nobel Prize in 939. Today, the largest accelerator in the U.S. is Fermilab (near Chicago), which is run by the Department of Energy. Several UNL faculty members do research at Fermilab. Fermilab's most powerful accelerator, the Tevatron, accelerates protons to an energy of,000 gigaelectronvolts (,000 GeV). The ring is a little less than two kilometers in diameter and the protons travel close to the speed of light.) 3 Why is it important to be able to make particles go very fast? Researchers had found that alpha particles could induce transmutation in light elements, and it was natural to wonder whether artificial transmutation could be induced in heavier elements. Breaking up a heavier nucleus required more massive and/or faster-moving particles. 3.4. NUCLEAR BINDING ENERGY Accelerators provided many more ways to study the nucleus. A fundamental principle in physics is that objects always want to be in their lowest energy state. A fundamental question about the nucleus was what held the nucleus together? Why did it take so much energy to split a nucleus? 3.4.. The Whole is Less than the Sum of its Parts. It seems as though you ought to be able calculate the mass of an atom of an element by adding up the masses of the protons and neutrons in the element (Electron masses are so small compared to nucleons that they can be neglected.) The rest mass the mass when standing still of a hydrogen atom (a proton plus an electron) is.00794 u and the rest mass of the neutron is.00866 u. The rest mass of a proton by itself is.0077 u. A deuteron, H, is a proton, a neutron and an electron, which is a hydrogen atom plus a neutron. The rest mass of a deuteron is.040 u. If you add up the masses of the hydrogen atom and the neutron that make up the deuteron: H =.00794 u neutron =.00866 u total =.0660 u The sum of the rest masses of the particles by themselves is 0.0050 u greater than the deuteron rest mass. 3.4.. E=mc. In 906, Einstein discovered a consequence of his theory of special relativity. Although he presented it as an afterthought, the idea that there is equivalence between mass and energy is embodied in what is probably the most famous physics equation: E where m is the mass of the object, c is the speed of light (c = 3.00 0 8 = mc (3.4.) m s ), and E is the energy. Even a small amount of mass is equivalent to an enormous amount of energy because c is a really, really big number. http://www.llnl.gov/llnl/history/eolawrence.html has some more biographical information on Lawrence and also a picture of the first cyclotron. 3 http://www.fnal.gov/ has pictures.
EXAMPLE 3.: What is the amount of energy contained in.00 g of matter? Draw a picture No picture is necessary known: m =.00 g =.00 0-3 kg need to find: E = energy Equation to use E = mc ( 3 )( 8 m ) s Plug in numbers E =.00 0 kg 3.00 0 3 = 9.00 0 J Answer: Comment: 3 E = 9.00 0 J This is equivalent to roughly 0 tons of TNT, or the amount of energy consumed in the whole United States on average every 30 seconds. The amount of mass the nucleons lose when they form a nucleus is called the mass defect. The mass defect is the difference between the mass of the protons and neutrons separately and the mass of the nucleus. The energy lost when the protons and neutrons bind together to form a nucleus is called the binding energy. The binding energy tells you how much energy you get out when the individual nucleons form a nucleus. The binding energy is given by BE = mass of parts - mass of atom (3.4.) where the mass of the parts is the mass of the protons, neutrons and electrons and the mass of the particular isotope. Remember that the mass given on the periodic table is an average of the masses of all the naturally occurring isotopes, so you have to have the mass of the specific isotope you are studying. In general, this will be given to you. Figure 3.3 is a graph of binding energy per nucleon (i.e. the binding energy divided by the number of protons + neutrons) vs. mass number. The graph has a maximum near mass number 56 (which is iron). The most stable nucleus (which is the one with the highest binding energy per nucleon) is 56 6 Fe. This is an important dividing line because it determines the method by which the nucleus can best reach its lowest energy state. Nuclei to with a higher mass number tend to release energy by fission (the topic of the next chapter) and nuclei with a lower mass number tend to release energy via fusion (the topic of the last chapter in this unit). It is energetically advantageous for a proton 0 0 40 60 80 00 0 40 60 80 00 0 40 Mass number (A) and a neutron to form a nucleus than it is for them to exist separately because their energy is lower when they are combined. The inverse situation is that if you want to separate the proton and the Binding energy per nucleon (MeV) 9 8 7 6 5 4 3 Energy released by fusion Energy released by fission Figure 3.3: Binding energy per nucleon vs. mass number 4 4 http://resourcefulphysics.org/freematerial/advanced/text/isotopes_and_the_unstable_nucleus.doc
neutron once they are in the nucleus, you have to add energy to overcome the preference for being a nucleus and not independent particles. The higher the binding energy per nucleon, the more energy necessary to take the nucleus apart. He- 4 has a high binding energy per nucleon compared with deuterium. If we have two deuterium nuclei and a He-nucleus, each would have four protons and four neutrons. If there were a way to put together the two deuterium nuclei to form a He-4 nucleus, there would be a lot of excess energy left over. The binding energy for Fe-56 is 8.8 MeV/nucleon. This means that if you could separate an iron nucleus into its protons and neutrons, you would get (6*8.8 MeV = 30 MeV) of energy released. 3.4.3. Application to the Deuteron. We found that a deuteron looses 0.00305 u of mass when the individual particles come together to form the nucleus. This change in mass means that energy is released when individual nucleons form the deuteron. Equation (3.4.) can be used to calculate how much energy is released. The conversion factor to covert from atomic mass units u to kg is u =.66 0-7 kg. EXAMPLE 3.: How much energy is released when a deuteron is formed? neutron H atom deuteron Draw a picture n 0 H H m =.008665 u m =.00794 u m =.0355 u known: m = mass lost = 0.0050 u need to find: E = energy Equation to use E = mc We need to convert the mass into kg -7.66 0 kg -30 0.0050 u =4.500 0 kg u E = mc -30 8 Plug in numbers m =(4.500 0 kg) ( 3.00 0 s ) -3 Answer: (3 s.f.) E = 3.74 0 J -3 =3.735 0 J Comment: See following discussion The amount of energy released seems like a very small number; however, remember that this is the energy lost for a single deuteron. If we have a mole of deuterons ( mole = 6.0 0 3 deuterons), then the total energy would be: E 3-3 J 6.0 0 deuterons = 3.735 0 =.5 0 mole deuteron J mole
EXAMPLE 3.3: 3.735 0-3 J of energy are released when a deuteron forms. How much energy is this in units of ev? neutron H atom deuteron Draw a picture n 0 H H m =.008665 u m =.00794 u m =.0355 u known: E = energy lost = 3.735 0-3 J need to find: E = energy lost in ev Equation to use This is a conversion. The conversion factor we need to use is: -9 ev =.60 0 J Plug in numbers Answer: (3 s.f.) Comment: E 6 E =.33 0eV -3 = 3.735 0 J.60 0-9 J 6 =.334375 0 ev ev We also could write this in MeV. The energy released in forming a deuteron is.33 MeV. This can be confirmed by an experiment. Deuterium can be formed when a hydrogen atom is bombarded with neutrons. The neutron is captured and a gamma ray is emitted. The gamma ray carries to.3 MeV of energy that is released when nucleus is formed. n+ H H + γ 0 3.4.4. Implications for Radioactive Decay. Understanding the difference in energy and the desire of all objects to be in their lowest energy states helps to explain why some nuclei decay on their own. Radioactive decay is a way for an atom to move to a lower energy state. A nucleus will decay if the energy of the nucleus is higher than the energy of whatever the nucleus could decay into. U-38 spontaneously decays by emitting an alpha particle. U Th+ He 38 34 4 9 90 The fact that this happens spontaneously (i.e. without needing to hit it with another particle) suggests that the energy of the two nuclei on the right-hand side is lower than the energy of the U-38. The mass of U-38 is 38.05078 u, the mass of Th-34 is 34.03660 u and the mass of the alpha particle is 4.005 u. The mass on the left-hand side of the equation is 38.05079 u and the mass on the right-hand side of the equation is 38.038 u. Energy is released in the decay. EXAMPLE 3.4: How much energy is released when a mole of U-38 decays by alpha decay? Draw a picture known: No picture needed m L = mass of one mole of the lefthand side = 38.05078 g m R = mass of one mole of the right-hand side = 38.038 g need to find: E = energy lost by one mole of U-38 decaying
The total mass lost is 38.05078 g 38.038 g = 0.067 g =.67 0-5 kg. Equation to use E = mc Plug in numbers E = (.37 0-5 kg)( 3.00 0 8 m ) s =.33 0 J Answer: (3 s.f.) Comment: E =. 0 J This is about the amount of energy that results from burning around 4000 gallons of gasoline (See Table III.). 3.5. NUCLEAR STABILITY When we introduced radioactivity, you were told whether a nucleus would decay and, if it did, what type of particle would be emitted. The key is, as usual, energy and the desire of almost everything to get in the lowest energy state possible. Nuclides is a general term that describes elements and their isotopes. Each dot on Figure 3.4 is a nuclide that has been arranged on a graph of number of neutrons vs. number of protons. The stable nuclides (the ones that do not spontaneously decay) fall within a narrow band, known as the band of stability. Although there are a few non-stable nuclei within the band, the majority of the nuclides in this band are stable. Isotopes that lie outside the band (which stops at proton number = 83) are unstable. The band of stability lies on the neutronrich side of the line that describes number of protons equals number of neutrons (the line marked N = Z in Figure 3.4. As the number of protons in the nucleus Figure 3.4: Nuclei stability increases, the region of stability moves further away 5 from the N = Z line. This is because more neutrons are required to compensate for the larger proton-proton repulsions. Isotopes above and to the left of the band (and below Z = 83) tend to be beta emitters they want to lose a neutron and gain a proton so that they move closer to the band of stability. Isotopes below and to the right (again, up to Z = 83) of the band tend to be positron emitters because they prefer to lose a proton and gain a neutron. All isotopes with more than 83 protons tend to be alpha emitters, since they want to reduce the overall number of nucleons. There are a couple of special circumstances. Isotopes for which the number of protons and the number of neutrons both are even are more stable than isotopes for which the number of protons and the number of neutrons both are odd. There are 64 stable isotopes; only 5 have both odd numbers of protons and neutrons. 57 have even numbers of both and the rest are mixed even/odd. 6 Another observation is that when the protons and neutrons add up to certain numbers, the isotope is more 5 http://www.plusphysics.com/nucleus_and_radioactivity/study_material.asp?chapter=&page=3 6 This has to do with the pairing of the magnetic moments of the neutrons and protons.
stable. These numbers are called magic numbers and are, 8, 0, 8, 50, 8, and 6. 4 He, 6 8O, and 0Ca 40 are very stable because both the neutrons by themselves and the protons by themselves are magic numbers, plus they number of neutrons and the number of protons is even. 3.6. SUMMARIZE 3.6.. Definitions: Define the following in your own words. Write the symbol used to represent the quantity where appropriate.. Linear accelerator. Electron-volt (the unit) 3. Cyclotron accelerator 4. Rest mass 5. Deuteron 6. Mass defect 7. Binding energy 8. Nuclide 3.6.. Equations: For each question: a) Write the equation that relates to the quantity b) Define each variable by stating what the variable stands for and the units in which it should be expressed, and c) State whether there are any limitations on using the equation.. The equation relating mass and energy. 3.6.3. Concepts: Answer the following briefly in your own words.. Explain how a linear accelerator produces a beam of fast-moving charged particles.. How does the chart of nuclear stability tell you whether an isotope is likely to be a beta emitter, a positron emitter or an alpha emitter? 3. The isotope 6 He is unstable. What particle do you expect it to emit? 4. What is a stable nuclide? 5. When protons and neutrons join together to make a nucleus (fusion), energy is a) released b) absorbed
c) unchanged d) could do either depending on the specific nuclei 6. Which one of the following is most likely to experience fission? Explain the rationale behind your selection. a) Iridium b) Iron c) Hydrogen d) Helium 7. The vast majority of atoms we come into contact with are stable. Given that most known isotopes are unstable, how can this be? 8. Explain in your own words what binding energy is. 9. What was the problem with the linear accelerator that spurred Lawrence to create the cyclotron? 0. Explain what binding energy per nucleon means 3.6.4. Your Understanding. What are the three most important points in this chapter?. Write three questions you have about the material in this chapter. 3.6.5. Questions to Think About. Why are protons more effective projectiles for producing nuclear reactions than are α particles or heavy ions? 3.6.6. Problems. How much energy is contained the mass of a 80.0-kg person? How many tons of TNT is that amount of energy equivalent to?. The isotope 4 Mg has a mass of 3.98504 u. What is the mass defect? What is the binding energy for this isotope? 3. How much energy must be supplied to break a single iron-56 nucleus into separate protons and neutrons? The mass of an iron-56 nucleus is 55.906 u. 4. How much energy must be supplied to break a single lithium-7 nucleus into separate protons and neutrons? The mass of a Li-7 nucleus is 7.0435 u. 5. The mass of an electron is 5.485799 0-4 u. What is the energy associated with one electron? 6. What is the mass and energy loss when two protons and two neutrons combine to make a He-4 nucleus. The rest mass of the He-4 nucleus is 4.0050 u. 7. Calculate the sum of the masses for the nucleons in a helium nucleus. Is there a mass defect? How much is it? 8. What is the energy in ev that is released when one mole of deuterons is formed?
9. If you hit a deuteron with a γ particle, you can form a neutron and a proton if the γ has enough energy: H + γ n+ H 0 0. How much energy does the γ need to break up the deuteron into a neutron and a proton?
HW Covers Class 3 and is due April nd, 007 PHYS 6 Spring 007 HW 33. The vast majority of atoms we come into contact with are stable. Given that most known isotopes are unstable, how can this be?. ( points) The isotope 4 Mg has a mass of 3.98504 u. a) What is the mass defect? b) What is the binding energy for this isotope?