Physics (5th Edition)
5th Edition
ISBN: 9780321976444
Author: James S. Walker
Publisher: PEARSON
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Chapter 31, Problem 5PCE
To determine
The distance of closest approach between the alpha particle and the gold nucleus for the case
K = 5.3 MeV
.
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Chapter 31 Solutions
Physics (5th Edition)
Ch. 31.1 - Prob. 1EYUCh. 31.2 - Prob. 2EYUCh. 31.3 - Prob. 3EYUCh. 31.4 - Prob. 4EYUCh. 31.5 - Prob. 5EYUCh. 31.6 - Prob. 6EYUCh. 31.7 - Prob. 7EYUCh. 31 - Prob. 1CQCh. 31 - Prob. 2CQCh. 31 - Prob. 3CQ
Ch. 31 - Prob. 4CQCh. 31 - Prob. 5CQCh. 31 - Prob. 6CQCh. 31 - Prob. 7CQCh. 31 - Prob. 8CQCh. 31 - Prob. 9CQCh. 31 - Prob. 1PCECh. 31 - Prob. 2PCECh. 31 - Prob. 3PCECh. 31 - Prob. 4PCECh. 31 - Prob. 5PCECh. 31 - Prob. 6PCECh. 31 - Prob. 7PCECh. 31 - Prob. 8PCECh. 31 - Prob. 9PCECh. 31 - Prob. 10PCECh. 31 - Prob. 11PCECh. 31 - Prob. 12PCECh. 31 - Prob. 13PCECh. 31 - Prob. 14PCECh. 31 - Prob. 15PCECh. 31 - Prob. 16PCECh. 31 - Prob. 17PCECh. 31 - Prob. 18PCECh. 31 - Prob. 19PCECh. 31 - Prob. 20PCECh. 31 - Prob. 21PCECh. 31 - Prob. 22PCECh. 31 - Prob. 23PCECh. 31 - Prob. 24PCECh. 31 - Prob. 25PCECh. 31 - Prob. 26PCECh. 31 - Prob. 27PCECh. 31 - Prob. 28PCECh. 31 - Prob. 29PCECh. 31 - Prob. 30PCECh. 31 - Prob. 31PCECh. 31 - Prob. 32PCECh. 31 - Prob. 33PCECh. 31 - Prob. 34PCECh. 31 - Prob. 35PCECh. 31 - Prob. 36PCECh. 31 - Prob. 37PCECh. 31 - Prob. 38PCECh. 31 - Prob. 39PCECh. 31 - Prob. 40PCECh. 31 - Prob. 41PCECh. 31 - Prob. 42PCECh. 31 - Prob. 43PCECh. 31 - Prob. 44PCECh. 31 - Prob. 45PCECh. 31 - Prob. 46PCECh. 31 - Prob. 47PCECh. 31 - Prob. 48PCECh. 31 - Prob. 49PCECh. 31 - Prob. 50PCECh. 31 - Prob. 51PCECh. 31 - Prob. 52PCECh. 31 - Give the electronic configuration for the ground...Ch. 31 - Prob. 54PCECh. 31 - Prob. 55PCECh. 31 - Prob. 56PCECh. 31 - The configuration of the outer electrons in Ni is...Ch. 31 - Prob. 58PCECh. 31 - Prob. 59PCECh. 31 - Prob. 60PCECh. 31 - Prob. 61PCECh. 31 - Prob. 62PCECh. 31 - Prob. 63PCECh. 31 - Prob. 64PCECh. 31 - Prob. 65PCECh. 31 - Prob. 66PCECh. 31 - Prob. 67PCECh. 31 - Prob. 68GPCh. 31 - Prob. 69GPCh. 31 - Prob. 70GPCh. 31 - Prob. 71GPCh. 31 - Prob. 72GPCh. 31 - Prob. 73GPCh. 31 - Prob. 74GPCh. 31 - Prob. 75GPCh. 31 - Prob. 76GPCh. 31 - Prob. 77GPCh. 31 - Prob. 78GPCh. 31 - Prob. 79GPCh. 31 - Prob. 80GPCh. 31 - Prob. 81GPCh. 31 - Prob. 82GPCh. 31 - Prob. 83GPCh. 31 - Prob. 84PPCh. 31 - Prob. 85PPCh. 31 - Prob. 86PPCh. 31 - Prob. 87PPCh. 31 - Prob. 88PPCh. 31 - Prob. 89PP
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- An 8.3 MeV alpha particle is shot directly toward the nucleus of a gold atom (atomic number 79). What is the distance of closest approach of the alpha particle to the nucleus?arrow_forwardEach a particle in a beam of a particles has a kinetic energy of 4.5 MeV. Through what potential difference would you have to accelerate these a particles in order that they would have enough energy so that if one is fired head-on at a gold nucleus it could reach a point 1.9 10^-14 m from the center of the nucleus?arrow_forwardIn a scattering experiment, an alpha particle A is projected with the velocity up = -(600 m/s)i + (750 m/s)j - (800 m/s)k into a stream of oxygen nuclei moving with a common velocity vo = (630 m/s)j. After colliding successively with nuclei B and C, particle A is observed to move along the path defined by the Points A₁(280, 240, 120) and A2(360, 320, 160), while nuclei B and Care observed to move along paths defined, respectively, by B₁(147, 220, 130), B2(114, 290, 120), and by C₁(240, 232, 90) and C₂(240, 280, 75). All paths are along straight lines and all coordinates are expressed in millimeters. Knowing that the mass of an oxygen nucleus is four times that of an alpha particle, determine the speed of each of the three particles after the collisions. The speed of particle A is The speed of particle B is The speed of particle Cis m/s. m/s. m/s.arrow_forward
- In an alpha particle (42He) scattering experiment, using a thin gold (19779Au) foil, the initial kinetic energy of the alpha particle is 2.0MeV. What is the potential energy of the alpha particle/gold nucleus system at closest impact?arrow_forwardIn a Rutherford scattering experiment, an a-particle (charge = +2e) heads directly toward a gold nucleus (charge = +79e). The α-particle had a kinetic energy of 5.0 MeV when very far (r→ ∞) from the nucleus. Assuming the gold nucleus to be fixed in space, determine the distance of closest approach. Hint: Use conservation of energy with PE =kq1q2/r.arrow_forwardElectron capture is a variant on beta-radiation. The lightest nucleus to decay by electron capture is 7Be -- beryllium-7. The daughter nucleus is 7Li -- lithium-7. The electron is transformed into a massless particle (a neutrino): e +"Be* → "Li + v The initial electron is bound in the atom, so the beryllium mass includes the electron. Infact, since the electron starts bound in the atom, a more-accurate statement of the nuclear reaction is probably: "Be → "Li + v The masses are beryllium: 7.016929 u, and lithium: 7.016003 u, and refer to the neutral atom as a whole. (Use uc and uc² as your momentum and energy units -- but carry them along in your calculation.) The initial beryllium atom is stationary. Calculate the speed of the final lithium nucleus in km/s. (You will make life much easier for yourself if you recognize that practically all the energy released goes into the lighter particle. c = 300,000 km/s)arrow_forward
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