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Homework Assignment 7 Physics 55 Made available: Due in class: Tuesday, November 1, 2005 Monday, November 7, 2005 Problem 1: Problems from the Text 1. Do problem 21 on page 519 of the text. 2. Do problem 21 on page 541. 3. Do problem 24 on page 542 of the text. 4. Do problem 29 on page 542 of the text. Problem 2: Some Problems Not in the Textbook 1. Some practice with nuclear reactions. (a) The radioactive phosphorus nucleus trino ν̄ 32 15 P beta decays by emitting an electron e and an antineu- 32 15 P 0 → xy X + −1 e + 00 ν̄, (1) where ν is the lower-case Greek letter nu. An anti-neutrino is a neutral particle of negligible mass but that can have a kinetic energy. i. What is the atomic number y, the atomic mass number x, and the element X that results from this decay? (You will likely need access to a periodic table of elements to determine the element X.) ii. If the maximum energy of the emitted electron is 1.7 MeV, what is the atomic mass of the daughter nucleus X in atomic mass units u? Data and a hint: An MeV (million electron volts) is a widely used unit of energy for nuclear and particle reactions, with 1 MeV ≈ 1.6 × 10−13 J. An atomic mass unit u is a widely used unit of mass for nuclei, with 1 u ≈ 1.66 × 10−27 kg, about the mass of one proton. From a table of isotopes, one can look up that the mass of a 32 15 P nucleus is 31.973907 u. Finally, use conservation of mass-energy to deduce the mass of the daughter nucleus X: total mass-energy before the reaction (simply the rest energy mc2 of the phosphorus nucleus) has to equal the total mass-energy after (the rest energy of mystery element X plus the kinetic energy of the electron). For this problem, you can assume that the nucleus X has zero speed after the decay (it is so massive compared to an electron that its speed is negligible) and you can ignore the kinetic energy of the antineutrino, which is zero when the emitted electron has its maximum kinetic energy. (b) As we will see in Chapter 17 of the text, in a later stage of evolution, some stars begin to fuse two 126 C nuclei (each of mass 12.000000 u) into one 24 12 Mg nucleus (which has a mass of 23.985042 u). i. In MeV energy units, how much energy is released by this fusion reaction? ii. You can figure out what the minimum absolute temperature T the core of the star must have for this carbon fusion reaction to occur by calculating the minimum speed two +6-charged carbon nuclei must have to approach each other such that the nuclei are able to overcome their enormous electrical repulsion and just touch each other, in which case the short-ranged strong 1 interaction becomes involved and a nuclear reaction can occur. This speed then corresponds to a certain large temperature. The first step is for you to calculate the electrical (also called Coulombic) potential energy of two carbon nuclei in MeV units when the two nuclei are just touching each other. (This is the amount of energy that will be released when the two nuclei are allowed to move away from each other a large distance.) You can calculate the electrical potential energy by knowing that the radius of a carbon nucleus is about 3 × 10−15 m, that the charge Q = 6e of a carbon nucleus is six times the charge e of a proton (with value e ≈ 1.6 × 10−19 C, where the letter C denotes a coulomb of charge), and that the formula for the electrical potential energy of two particles with charges Q1 and Q2 that are a distance d apart is given by the equation Eelectrical = 9.0 × 109 Q1 Q2 , d (2) where the charges Q1 and Q2 are measured in units of coulombs, the distance d is measured in units of meters m, and the energy Eelectrical has units of joules J. iii. Each carbon nucleus needs to have an initial kinetic energy that is at least half of the above electrical potential energy, if they are to overcome the electrical repulsion and approach within a radius of each other. The kinetic theory of gases relates the average kinetic energy K of a molecule in a gas of temperature T by the formula K= 3 kT, 2 (3) where the the letter k ≈ 1.4×10−23 J/K is the so-called Boltzmann constant. Use this equation to deduce the minimum temperature that the core of a star must have for carbon-carbon fusion reactions to occur, giving your answer as a multiple of the Sun’s core temperature of 15,000,000 K. 2. (a) Assuming that the luminosity and the rate of hydrogen fusion into helium in the Sun remains constant throughout the lifetime of the Sun, determine what fraction of the Sun’s mass will be converted into helium over the remaining 5 billion years of the Sun’s life. (b) About how many Earths does this fraction of the Sun’s mass correspond to? (When writing science articles for the general public, authors like to relate some big number to some image that is easily appreciated by the reader.) (c) Given that the current percentage composition (by mass) of the Sun is 70% hydrogen, 28% helium, and 2% heavier elements, determine the composition of the Sun after 5 billion more years of fusion. 3. The star Sirius has a luminosity that is 23.5 times the luminosity of the Sun and has a mass that is 2.3 times the mass of the Sun. Explain whether the lifetime of Sirius will be longer, shorter, or about the same as the lifetime of the Sun. 4. The visual binary 70 Ophiuchi has a period of 87.7 years. The parallax angle of 70 Ophiuchi is 0.2” and the apparent length of the semimajor axis as seen through a telescope is 4.5”. (a) What is the distance from Earth to 70 Ophiuchi in light years? (b) What is the length of the semimajor axis in AU? (c) What is the sum of the masses of the two stars, in units of the mass of the Sun? 2 Problem 3: Extra Credit Problems 1. Use conservation of mass-energy and conservation of momentum to show that a proton p must have a kinetic energy of at least 3.23 MeV for the nuclear reaction 136 C(p, n)137 N to occur. Without taking conservation of momentum into account and just using conservation of mass-energy, one would conclude incorrectly that a smaller minimum kinetic energy of 3.00 MeV would suffice for this reaction to occur. Some data: the mass of a hydrogen atom is 1.007825 u, of the 136 C atom is 13.003355 u, of the nitrogen atom 137 N is 13.005739 u, and of the neutron is 1.08665. A similar issue arises in chemical reactions, a greater initial kinetic energy (called the threshold energy) is needed for an incoming atom than the amount based just on energy conservation for some reaction to occur because of the need to conserve momentum. 2. In lecture, I showed a picture of a bubble chamber in which an invisible photon (invisible since it was uncharged and so did not create a trail of bubbles) passed near some nucleus and then disappeared, producing an electron-positron pair in its place. Explain why a photon of arbitrarily high energy in vacuum can not simply convert into an electron-positron pair, the photon has to pass near some third massive particle like a nucleus. Hint: Think about this problem from a reference frame moving with the center of mass of the electronpositron pair after they have been created and deduce a contradiction. 3. Explain why it is impossible for an isolated free electron (actually, for any point particle with no internal structure) to absorb or emit a photon. Problem 4: Comments about the Homework and Course • About how long did it take you to complete this assignment? • Do you feel that you are understanding the course material? If not, please indicate what topics or ideas you would like to understand better. • Comments or suggestions about other parts of the course such as reading, homeworks, lectures, or observation sessions? 3