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# week 14 short assignments 1220

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This 17 page Bundle was uploaded by Dragon Note on Monday May 9, 2016. The Bundle belongs to 1220 at University of Missouri - Columbia taught by Y Zhang in Spring 2016. Since its upload, it has received 17 views. For similar materials see College Physics II in Physics 2 at University of Missouri - Columbia.

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Date Created: 05/09/16

SHort Assignment by 5/2/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... SHort Assignment by 5/2/2016 Due: 11:00am on Monday, May 2, 2016 To understand how points are awarded, read theing Policy for this assignment. The Basics of Nuclear Physics Learning Goal: To understand the notation and basic quantities involved in nuclear physics. Nuclear physics borrows the symbols for elements from chemistry. However, knowing which element we are dealing with only tells us one of the numbers important to nuclear physics. When referring to a specific nucleus, we use the following notation: . The superscript, 56 in this case, denotes the total number of nucleons (protons and neutrons) in the nucleus. This is called the nucleon number and is given the symbol . The subscript, 26 in this case, is the number of protons in the nucleus. This is called the atomic number and is given the symbol. An atom's atomic number determines which element the atom is, in this case (iron). Another important number characterizing an atom is the neutron number . Since is the total number of nucleons in the nucleus, the neutron number may be found from the equation . Solving for gives . Nuclei with the same atomic number but different neutron numbers are called isotopes. Isotopes are often written in a form such as "iron-56." Part A What is the atomic number of ? Express your answer as an integer. ANSWER: = 3 protons Answer Requested Part B What is the nucleon number of carbon-14? Express your answer as an integer. ANSWER: = 14 nucleons Answer Requested Part C What is the neutron number of ? Express your answer as an integer. ANSWER: SHort Assignment by 5/2/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... = 30 neutrons Answer Requested Part D Which of the following choices lists a pair of isotopes? ANSWER: and and and and Answer Requested It has been found that the radii of most nuclei are well approximated by the equation , where is an experimentally determined constant. Because the mass of a nucleon (a proton or a neutron) is close to one atomic mass unit (), the nucleon number is sometimes called the mass number. Part E What is the approximate radius of ? Express your answer in meters to two significant figures. Hint 1. What is ? Which of the following numbers correctly gives the valuforf ? ANSWER: 290 208 126 82 ANSWER: −15 = 7.10×10 SHort Assignment by 5/2/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... All attempts used; correct answer displayed Part F Assuming that each nucleus is roughly spherical and that its mass is roughly equal to(in atomic mass units ), what is the density of a nucleus with nucleon number ? Express your answer in terms of , , and . Hint 1. Find the volume of the nucleus Recall that the volume of a sphere is given by . Assuming that the nucleus is spherical, find its volume. Express your answer in terms of , , and . ANSWER: = ANSWER: = All attempts used; correct answer displayed The fact that does not appear in your result is a very important property of nuclei. The observation that most nuclei have about the same density was key to the development of George Gamow's liquid drop model of nuclear binding. Nuclear Binding Energy Some atomic masses Particle Symbol Mass ( ) Electron 0.00055 Proton 1.00728 Neutron 1.00866 Hydrogen 1.00783 Helium 4.00260 Part A Calculate the mass defect of the nitrogen nucleus . The mass of neutral is equal to 14.003074 atomic mass units. SHort Assignment by 5/2/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Express your answer in atomic mass units to four significant figures. Hint 1. Definition of mass defect The total mass of the individual nucleons that make up an atom is always greater than the total mass of the atom itself by an amount called the mass defect. The mass defect is equal to the binding energy (in units of atomic mass ), which is the amount of energy required to separate the nucleus into its constituent parts. ANSWER: 0.1124 All attempts used; correct answer displayed Part B Calculate the binding energy of the nitrogen nucleus . Express your answer in millions of electron volts to four significant figures. Hint 1. Definition of binding energy The binding energy of the atom is the amount of energy that is required to pull apart all the nucleons that make up the atom. It is basically the mass defect of the atom but in units of energy. One can determine the binding energy by calculating the difference between the rest mass energy of the atom's constituent particles and the rest mass energy of the atom. Hint 2. Conversion of mass to energy Recall that the conversion of rest mass to energy is given by Einstein's famous formula , where is the mass, is the speed of light, and is the rest mass energy. Hint 3. Converting joules to electron volts Conversion between joules and electron volts can be done using . ANSWER: = 104.7 All attempts used; correct answer displayed Part C Calculate the binding energy per nucleon of the nitrogen nucleus . Express your answer in millions of electron volts to three significant figures. SHort Assignment by 5/2/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... ANSWER: 7.48 All attempts used; correct answer displayed Part D Calculate the mass defect of the helium nucleus . The mass of neutral is given by . Express your answer in atomic mass units to three significant figures. Hint 1. Definition of mass defect The total mass of the individual nucleons that make up an atom is always greater than the total mass of the atom itself by an amount called the mass defect. The mass defect is equal to the binding energy (in units of atomic mass ), which is the amount of energy required to separate the nucleus into its constituent parts. ANSWER: −2 3.04×10 All attempts used; correct answer displayed Part E Calculate the binding energy of the helium nucleus . Express your answer in millions of electron volts to three significant figures. Hint 1. Definition of binding energy The binding energy of the atom is the amount of energy that is required to pull apart all the nucleons that make up the atom. It is basically the mass defect of the atom but in units of energy. One can determine the binding energy by calculating the difference between the rest mass energy of the atom's constituent particles and the rest mass energy of the atom. Hint 2. Conversion of mass to energy Recall that the conversion of rest mass to energy is given by Einstein's famous formula , where is the mass, is the speed of light, and is the rest mass energy. Hint 3. Converting joules to electron volts Conversion between joules and electron volts can be done using . SHort Assignment by 5/2/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... ANSWER: = 28.3 All attempts used; correct answer displayed Part F Calculate the binding energy per nucleon of the helium nucleus . Express your answer in millions of electron volts to two significant figures. ANSWER: 7.1 All attempts used; correct answer displayed Score Summary: Your score on this assignment is 0.0%. You received 0 out of a possible total of 4 points. Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Short Assignment by 5/4/2016 Due: 11:00am on Wednesday, May 4, 2016 To understand how points are awarded, read theding Policy for this assignment. Nuclear Decay Conceptual Question In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case. A periodic table of elements is available on the Tools page (linked to in the top right corner of the screen). Part A undergoes alpha decay. Determine the resulting nucleus. Enter your answer in the space below. For example, if the resulting nucleus is enter ^40_20Ca. Hint 1. Alpha decay An alpha decay is the emission of an alpha particle , which is a bound state of two protons and two neutrons. Thus, when a nucleus undergoes an alpha decay, its mass number decreases by 4 and its atomic number decreases by 2. This decay is expressed by the equation , where is the new nucleus created in the decay. ANSWER: All attempts used; correct answer displayed Part B undergoes beta-minus decay. Determine the resulting nucleus. Enter your answer in the space below. For example, if the resulting nucleus is enter ^40_20Ca. Hint 1. Beta-minus decay In a beta-minus decay, a neutron transforms into a proton, an electron , and an antineutrino . The electron and antineutrino are emitted from the nucleus. Since in beta-minus decays a neutron transforms into a proton, the mass number of a nucleus that undergoes a beta-minus decay doesn’t change, while its atomic number increases by 1. This decay is expressed by the equation , where is the new nucleus created in the decay. ANSWER: Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... All attempts used; correct answer displayed Part C undergoes beta-plus decay. Determine the resulting nucleus. Enter your answer in the space below. For example, if the resulting nucleus is enter ^40_20Ca. Hint 1. Beta-plus decay In a beta-plus decay, a proton transforms into a neutron, an antielectro, and neutrino . The antielectron and neutrino are emitted from the nucleus. Since in beta-plus decays a proton transforms into a neutron, the mass number of a nucleus that undergoes a beta-plus decay doesn’t change, while its atomic number decreases by 1. This decay is expressed by , where is the new nucleus created in the decay. ANSWER: All attempts used; correct answer displayed Part D undergoes gamma decay. Determine the resulting nucleus. Enter your answer in the space below. For example, if the resulting nucleus is enter ^40_20Ca. Hint 1. Gamma decay In a gamma decay, the nucleus drops from a higher energy state to a lower energy state in a way analogous to an electron changing energy levels in an atom. This decay involves the emission of one or more high-energy photons ( rays), but no change in the number and type of constituents in the nucleus. Since gamma decays are simply the release of photons, neither the mass number nor the atomic number of a nucleus that undergoes a gamma decay changes, as expressed by the equation . ANSWER: Typesetting math: 100% Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... All attempts used; correct answer displayed Radioactive Decay Conceptual Question A radioactive sample undergoes three different types of radioactive decay and emits three different types of particles. As shown in the figure, these particles are emitted into a region of space with a uniform magnetic field directed out of the plane of the figure. The particles follow the paths indicated, and none of them bends either into or out of the plane of the figure. In the following questions ignore the neutrinos and antineutrinos emitted in beta decays. Part A Which type of radioactive decay would produce a decay particle that would move along path A? Hint 1. Motion of charged particles in magnetic fields When a charged particle moves in a magnetic field, it is acted on by a magnetic force perpendicular to the direction of both the particle's velocity and the magnetic field. Thus the particle follows a curved path. Hint 2. The right-hand rule for magnetic force A charged particle moving through a region of magnetic field experiences a magnetic force perpendicular to both the velocity of the particle and the magnetic field vector at the point of interaction. The requirement that the force be perpendicular to both the velocity and magnetic field vectors specifies the direction of the force to within an algebraic sign. This algebraic sign can be determined by a "right-hand rule." For example, consider the situation illustrated below, where a positive charge moves with velocity in a region with a uniform magnetic field directed into the plane of the figure. Typesetting math: 100% Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... The magnetic force must be perpendicular to both and . Therefore, the force vector can only point either directly to the right or directly to the left. The right-hand rule specifies which of these two options is correct. To employ the right-hand rule: 1. Point the fingers of your right hand in the direction of the velocity vector. 2. Then rotate your hand until you can curl your fingers in the direction of the magnetic field. The direction of your thumb is the direction of the magnetic force on a positive charge. In the example here, your thumb should point to the left, so the magnetic force on the positive charge is directed to the left, as shown in the following figure. Note, however, that if the charge is negative, the direction of the magnetic force is opposite to that determined for a positive charge. Therefore in the example described here, if the charge were negative instead of positive, the magnetic force would be directed to the right. Hint 3. Radioactive decays Alpha decay is the emission of an alpha particle, which is a bound state of two protons and two neutrons. In beta-minus decay, a neutron transforms into a proton, an electron, and an antineutrino. The electron and antineutrino are emitted from the nucleus. In beta-plus decay, a proton transforms into a neutron, an antielectron, and a neutrino. The antielectron (also called a positron because of its positive charge) and neutrino are emitted from the nucleus. In gamma decay, the nucleus drops from a higher energy state to a lower energy state in a way analogous to an electron changing energy levels in an atom. This decay involves the emission of a high-energy photon, but no Typesetting math: 100% change in the number and type of constituents in the nucleus. Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... ANSWER: decay decay decay decay The type of decay cannot be determined. Correct Part B Which type of radioactive decay would produce a decay particle that would move along path B? Hint 1. Motion of charged particles in magnetic fields When a charged particle moves in a magnetic field, it is acted on by a magnetic force perpendicular to the direction of both the particle's velocity and the magnetic field. Thus the particle follows a curved path. Hint 2. The right-hand rule for magnetic force A charged particle moving through a region of magnetic field experiences a magnetic force perpendicular to both the velocity of the particle and the magnetic field vector at the point of interaction. The requirement that the force be perpendicular to both the velocity and magnetic field vectors specifies the direction of the force to within an algebraic sign. This algebraic sign can be determined by a "right-hand rule." For example, consider the situation illustrated below, where a positive charge moves with velocity in a region with a uniform magnetic field directed into the plane of the figure. The magnetic force must be perpendicular to both and . Therefore, the force vector can only point either directly to the right or directly to the left. The right-hand rule specifies which of these two options is correct. To employ the right-hand rule: 1. Point the fingers of your right hand in the direction of the velocity vector. Typesetting math: 100%ate your hand until you can curl your fingers in the direction of the magnetic field. Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... The direction of your thumb is the direction of the magnetic force on a positive charge. In the example here, your thumb should point to the left, so the magnetic force on the positive charge is directed to the left, as shown in the following figure. Note, however, that if the charge is negative, the direction of the magnetic force is opposite to that determined for a positive charge. Therefore in the example described here, if the charge were negative instead of positive, the magnetic force would be directed to the right. Hint 3. Radioactive decays Alpha decay is the emission of an alpha particle, which is a bound state of two protons and two neutrons. In beta-minus decay, a neutron transforms into a proton, an electron, and an antineutrino. The electron and antineutrino are emitted from the nucleus. In beta-plus decay, a proton transforms into a neutron, an antielectron, and a neutrino. The antielectron (also called a positron because of its positive charge) and neutrino are emitted from the nucleus. In gamma decay, the nucleus drops from a higher energy state to a lower energy state in a way analogous to an electron changing energy levels in an atom. This decay involves the emission of a high-energy photon, but no change in the number and type of constituents in the nucleus. ANSWER: decay decay decay decay The type of decay cannot be determined. Correct Part C Which type of radioactive decay would produce a decay particle that would move along path C? Typesetting math: 100% Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Hint 1. Motion of charged particles in magnetic fields When a charged particle moves in a magnetic field, it is acted on by a magnetic force perpendicular to the direction of both the particle's velocity and the magnetic field. Thus the particle follows a curved path. Hint 2. The right-hand rule for magnetic force A charged particle moving through a region of magnetic field experiences a magnetic force perpendicular to both the velocity of the particle and the magnetic field vector at the point of interaction. The requirement that the force be perpendicular to both the velocity and magnetic field vectors specifies the direction of the force to within an algebraic sign. This algebraic sign can be determined by a "right-hand rule." For example, consider the situation illustrated below, where a positive charge moves with velocity in a region with a uniform magnetic field directed into the plane of the figure. The magnetic force must be perpendicular to both and . Therefore, the force vector can only point either directly to the right or directly to the left. The right-hand rule specifies which of these two options is correct. To employ the right-hand rule: 1. Point the fingers of your right hand in the direction of the velocity vector. 2. Then rotate your hand until you can curl your fingers in the direction of the magnetic field. The direction of your thumb is the direction of the magnetic force on a positive charge. In the example here, your thumb should point to the left, so the magnetic force on the positive charge is directed to the left, as shown in the following figure. Typesetting math: 100% Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Note, however, that if the charge is negative, the direction of the magnetic force is opposite to that determined for a positive charge. Therefore in the example described here, if the charge were negative instead of positive, the magnetic force would be directed to the right. Hint 3. Radioactive decays Alpha decay is the emission of an alpha particle, which is a bound state of two protons and two neutrons. In beta-minus decay, a neutron transforms into a proton, an electron, and an antineutrino. The electron and antineutrino are emitted from the nucleus. In beta-plus decay, a proton transforms into a neutron, an antielectron, and a neutrino. The antielectron (also called a positron because of its positive charge) and neutrino are emitted from the nucleus. In gamma decay, the nucleus drops from a higher energy state to a lower energy state in a way analogous to an electron changing energy levels in an atom. This decay involves the emission of a high-energy photon, but no change in the number and type of constituents in the nucleus. ANSWER: decay decay decay decay The type of decay cannot be determined. Correct The path of the particle in the magnetic field indicates that the particle is positive. In both beta-plus decays and alpha decays a positive particle is emitted; therefore in this case the specific type of decay cannot be determined. The only way that you could tell the two apart is if you had some knowledge of the speed of the particles. With the speed and the curvature of the trajectories, you could determine the mass. However, you do not have any information about the speed (or the energy, from which you could determine the speed). A Decay Chain Typesetting math: 100% Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... When the daughter nucleus produced in a radioactive decay is itself unstable, it will eventually decay and form its own daughter nucleus. If the newly formed daughter nucleus is also unstable, another decay will occur, and the process will continue until a nonradioactive nucleus is formed. Such a series of radioactive decays is called a decay chain. A good example of a decay chain is provided by , a naturally occurring isotope of thorium. Part A The first step in the decay chain of is an alpha decay. What is the daughter nucleus formed by this first decay? Hint 1. Alpha decay In alpha-particle decay, a helium nucleus (consisting of two protons and two neutrons) is emitted from the parent atom, leaving it with four fewer nucleons and an atomic number reduced by two. Hint 2. Find the numbers of protons and neutrons in the daughter nucleus Recall that in radioactive decays the atomic number and mass number are conserved. How many protons and neutrons are there in the daughter nucleus when undergoes an alpha decay? ANSWER: 232 neutrons and 88 protons 88 neutrons and 232 protons 88 neutrons and 140 protons 140 neutrons and 88 protons ANSWER: Correct Part B What is the energy released in the first step of the thorium-232 decay chain? The atomic mass of is 232.038054 and the atomic mass of is 228.0301069 . Express your answer numerically in megaelectron volts. Typesetting math: 100%pproach the problem Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... The energy released in a decay depends on the difference in mass, , of the system before and after the decay. It can be calculated using the relation . You need to take into account the mass of the particle emitted in the decay when calculating the mass of the system after the decay. Also, note that the number of electrons in the parent nucleus is the same as the number of electrons in the daughter nucleus plus the number of electrons in the alpha particle, so the electrons do not contribute to the total mass difference. Hint 2. Find the mass difference What is the difference in mass, , of the system before and after the alpha decay? Express your answer numerically in atomic mass units to four significant digits. Hint 1. Atomic mass of an alpha particle The atomic mass of an alpha particle is 4.002603. ANSWER: = 0.005343 ANSWER: = 4.98 Correct Part C then decays through a series of beta-minus decays; eventually, another isotope of thorium, , is formed. How many beta-minus decays will occur during this chain process? Hint 1. Beta decay In a beta-minus decay, an electron (as well as an antineutrino with negligible mass) is emitted as a neutron is converted into a proton. Such a decay increases the atomic number by one without changing the number of nucleons. ANSWER: 1 2 4 6 Correct Typesetting math: 100% Short Assignment by 5/4/2016 https://session.masteringphysics.com/myct/assignmentPrintView?displ... Part D then decays by emitting alpha particles until is formed. How many alpha decays will occur during this chain process? Hint 1. Alpha decay In alpha-particle decay, a helium nucleus (consisting of two protons and two neutrons) is emitted from the parent atom, leaving it with four fewer nucleons and an atomic number reduced by two. ANSWER: 2 4 6 8 Correct The chain continues with two more intermediate states until the stable isotope of lead is formed, after a total of 10 decays. Score Summary: Your score on this assignment is 58.3%. You received 1.75 out of a possible total of 3 points. Typesetting math: 100%

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