Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 44, Problem 37AP
To determine
The order of magnitude of the amount of neutrinos passing through the mother.
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A standard nuclear power plant generates 3.0 GWGW of thermal power from the fission of 235U235U. Experiments show that, on average, 0.19 uu of mass is lost in each fission of a 235U235U nucleus.
How many kilograms of 235U235U undergo fission each year in this power plant?
Express your answer in kilograms per year.
The half-lives of 235U and 238U are 7.04 × 108 years and 4.47 × 109 years, respectively, and the present abundance ratio is 238U>235U 5 137.7. It is thought that their abundance ratio was 1 at some time before our earth and solar system were formed about 4.5 × 109 years ago. Estimate how long ago the supernova occurred that supposedly produced all the uranium isotopes in equal abundance, including the two longest lived isotopes, 238U and 235U.
How much energy (in x 1016 Joule) does the Sun burn 1 kg of hydrogen fuel in a nuclear reaction?
Chapter 44 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 44.2 - Prob. 44.1QQCh. 44.5 - Prob. 44.3QQCh. 44.5 - Prob. 44.4QQCh. 44.8 - Prob. 44.5QQCh. 44.8 - Prob. 44.6QQCh. 44 - Prob. 1PCh. 44 - Prob. 2PCh. 44 - Prob. 3PCh. 44 - Prob. 4PCh. 44 - Prob. 5P
Ch. 44 - Prob. 6PCh. 44 - Prob. 7PCh. 44 - Prob. 8PCh. 44 - Prob. 9PCh. 44 - Prob. 10PCh. 44 - Prob. 11PCh. 44 - Prob. 12PCh. 44 - Prob. 13PCh. 44 - Prob. 14PCh. 44 - Prob. 15PCh. 44 - Prob. 16PCh. 44 - Prob. 17PCh. 44 - Prob. 18PCh. 44 - Prob. 20PCh. 44 - Prob. 21PCh. 44 - Prob. 22PCh. 44 - Prob. 23PCh. 44 - Prob. 24PCh. 44 - Prob. 25PCh. 44 - Prob. 26PCh. 44 - Prob. 27PCh. 44 - Prob. 29PCh. 44 - Prob. 30PCh. 44 - The various spectral lines observed in the light...Ch. 44 - Prob. 33PCh. 44 - Prob. 34APCh. 44 - Prob. 35APCh. 44 - Prob. 36APCh. 44 - Prob. 37APCh. 44 - Prob. 38APCh. 44 - Prob. 39APCh. 44 - Prob. 40APCh. 44 - An unstable particle, initially at rest, decays...Ch. 44 - Prob. 42APCh. 44 - Prob. 43APCh. 44 - Prob. 44APCh. 44 - Prob. 45APCh. 44 - Prob. 46CPCh. 44 - Prob. 47CPCh. 44 - Prob. 48CPCh. 44 - Prob. 49CP
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- The power output of the Sun is 41026W. (a) If 90% of this is supplied by the protonproton cycle, how many protons are consumed per second? (b) How many neutrinos per second should there be per square meter at the Earth from this process? This huge number is indicative of how rarely a neutrino interacts, since large detectors observe very few per day.arrow_forwardThe primary decay mode for the negative pion is +v . (a) What is the energy release in MeV in this decay? (b) Using conservation of momentum, how much energy does each of the decay products receive, given the is at rest when it decays? You may assume the muon antineutrino is massless and has momentum p = E/c , just like a photon.arrow_forwardSuppose that a tau neutrino and a tau antineutrino, both of which are just barely moving, encounter each other in space and completely annihilate to form two photons of equal energy. In view of the uncertainty about the mass of the tau neutrino (<18.2 MeV/c2), what is the shortest wavelength lamda0 of light that could be emitted by the annihilation? lamda0 = ? m Would the light be visible to the human eye?arrow_forward
- The primary decay mode for the negative pion is π− → μ− + ν-μ . (a) What is the energy release in MeV in this decay? (b) Using conservation of momentum, how much energy does each of the decay products receive, given the π− is at rest when it decays? You may assume the muon antineutrino is massless and has momentum p = E / c , justlike a photon.arrow_forwardThe Sun and all Main Sequence stars derive their luminosity by fusing hydrogen to helium; in particular 4 1H ® 1 4He + Energy (photons) The mass of a proton is 1.0078 AMU, while the mass of a helium nucleus is 4.0026 AMU. a) How much mass is ``lost'' (in AMU and kg) in fusing four H atoms to one helium? This is called the mass defect. b) Mass isn't really lost, of course. It is converted to energy via E=mc2. How much energy in joules (J) is liberated in a single reaction? c) How many reactions per second are required to account for the entire luminosity of the Sun?arrow_forwardThe power output of the Sun is 4 × 1026 W. Part (a) If 88 % of this is supplied by the proton-proton cycle, how many protons are consumed per second? Part (b) How many neutrinos per second should be incident on every square meter of the Earth from this process? This huge number is indicative of how rarely a neutrino interacts, since large detectors observe very few per day.arrow_forward
- In a fission process in which 235U235U is struck by a neutron and undergoes fission to produce144Ba, 89Kr, and three neutrons. The measured masses of these isotopes are 235.043930 u (235U), 143.922953 u (144Ba), 88.917630 u (89Kr), and 1.0086649 u (neutron). A) Calculate the energy (in MeV) released by each fission reaction. Express your answer in megaelectronvolts. B) Calculate the energy released per gram of 235U, in MeV/g. Express your answer in megaelectronvolts per gram.arrow_forwardThe power output of the Sun is 4×10^26 W. (a) If 90% of this energy is supplied by the proton-proton chain, how many protons are consumed per second? (b) How many neutrinos per second should there be per square meter at the surface of Earth from this process? Only part (b) is needed.arrow_forwardObservations of neutrinos emitted by the supernova SN1987a place an upper limit of 20 eV on the rest energy of the electron neutrino. If the rest energy of the electron neutrino were, in fact, 20 eV, what would be the speed difference between light and a 1.5 MeV electron neutrino?arrow_forward
- The Sun generates energy by p-p proton-proton chain nuclear fusion. The second step of p-p chain is: 2H + 1H → 3He. The particle masses are 2.0141 u, 1.0078 u, 3.0160 u, for 2H, 1H, 3He, respectively. This step produces _____ MeV of energy.arrow_forwardWhich of the following are possible reactions? (Select all that apply.) 1)10n + 23592U → 14054Xe + 9438Sr + 2(10n) 2) 10n + 23592U → 13250Sn + 10142Mo + 3(10n) 3)10n + 23994Pu → 12753I + 9341Nb + 3(10n)arrow_forwardCalculate the binding energy (in MeV) of the last neutron in the 4320Ca nucleus. (Hint: You should compare the mass of 4320Ca with the mass of 4220Ca plus the mass of the last neutron. The mass of 4220Ca = 41.958622 u, whereas the mass of 4320Ca = 42.958770 u.) MeVarrow_forward
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