Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 21, Problem 85P
To determine
The center to center separation of the alpha particle and the gold nucleus.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
In an x-ray tube, high-speed electrons are slammed into a lead target, giving off x-rays. If the electrons are accelerated from rest through a potential difference of 190 000 volts, what speed do they have when they strike the target? ( q e = 1.6 × 10 −19 C, m e = 9.11 × 10 −31 kg, and c = 3.00 × 10 8 m/s)
Plz help
In the Rutherford model of the hydrogen atom, a proton (mass M, charge Q) is the nucleus and an electron (mass m, charge q) moves
around the proton in a circle of radius r. Let k denote the Coulomb force constant (1/4TTE0) and G the universal gravitational constant.
The ratio of the electrostatic force to the gravitational force between electron and proton is:
O kMm/GQq
O kQq/GMm
O GQq/kMm
O GMm/kQq
O kQq/GMmr²
Chapter 21 Solutions
Physics for Scientists and Engineers
Ch. 21 - Prob. 1PCh. 21 - Prob. 2PCh. 21 - Prob. 3PCh. 21 - Prob. 4PCh. 21 - Prob. 5PCh. 21 - Prob. 6PCh. 21 - Prob. 7PCh. 21 - Prob. 8PCh. 21 - Prob. 9PCh. 21 - Prob. 10P
Ch. 21 - Prob. 11PCh. 21 - Prob. 12PCh. 21 - Prob. 13PCh. 21 - Prob. 14PCh. 21 - Prob. 15PCh. 21 - Prob. 16PCh. 21 - Prob. 17PCh. 21 - Prob. 18PCh. 21 - Prob. 19PCh. 21 - Prob. 20PCh. 21 - Prob. 21PCh. 21 - Prob. 22PCh. 21 - Prob. 23PCh. 21 - Prob. 24PCh. 21 - Prob. 25PCh. 21 - Prob. 26PCh. 21 - Prob. 27PCh. 21 - Prob. 28PCh. 21 - Prob. 29PCh. 21 - Prob. 30PCh. 21 - Prob. 31PCh. 21 - Prob. 32PCh. 21 - Prob. 33PCh. 21 - Prob. 34PCh. 21 - Prob. 35PCh. 21 - Prob. 36PCh. 21 - Prob. 37PCh. 21 - Prob. 38PCh. 21 - Prob. 39PCh. 21 - Prob. 40PCh. 21 - Prob. 41PCh. 21 - Prob. 42PCh. 21 - Prob. 43PCh. 21 - Prob. 44PCh. 21 - Prob. 45PCh. 21 - Prob. 46PCh. 21 - Prob. 47PCh. 21 - Prob. 48PCh. 21 - Prob. 49PCh. 21 - Prob. 50PCh. 21 - Prob. 51PCh. 21 - Prob. 52PCh. 21 - Prob. 53PCh. 21 - Prob. 54PCh. 21 - Prob. 55PCh. 21 - Prob. 56PCh. 21 - Prob. 57PCh. 21 - Prob. 58PCh. 21 - Prob. 59PCh. 21 - Prob. 60PCh. 21 - Prob. 61PCh. 21 - Prob. 62PCh. 21 - Prob. 63PCh. 21 - Prob. 64PCh. 21 - Prob. 65PCh. 21 - Prob. 66PCh. 21 - Prob. 67PCh. 21 - Prob. 68PCh. 21 - Prob. 69PCh. 21 - Prob. 70PCh. 21 - Prob. 71PCh. 21 - Prob. 72PCh. 21 - Prob. 73PCh. 21 - Prob. 74PCh. 21 - Prob. 75PCh. 21 - Prob. 76PCh. 21 - Prob. 77PCh. 21 - Prob. 78PCh. 21 - Prob. 79PCh. 21 - Prob. 80PCh. 21 - Prob. 81PCh. 21 - Prob. 82PCh. 21 - Prob. 83PCh. 21 - Prob. 84PCh. 21 - Prob. 85PCh. 21 - Prob. 86PCh. 21 - Prob. 87P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- Alpha decay is nuclear decay in which a helium nucleus is emitted. If the helium nucleus has a mass of 6.801027 kg and is given 5.00 MeV of kinetic energy, what is its velocity?arrow_forwardUnreasonable Results A proton has a mass of 1.671027 kg. A physicist measures the proton's total energy to be 50.0 MeV. (a) What is the proton's kinetic energy? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?arrow_forwardThe creation and study of new and very massive elementary particles is an important part of contemporary physics. To create a particle of mass M requires an energy Mc2 . With enough energy, an exotic particle can be created by allowing a fast-moving proton to collide with a similar target particle. Consider a perfectly inelastic collision between two protons: an incident proton with mass kinetic energy K, and momentum magnitude p joins with an originally stationary target proton to form a single product particle of mass M. Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle. (a) Show that the energy available to create a product particle is given by Mc2=2mpc21+K2mpc2 This result shows that when the kinetic energy K of the incident proton is large compared with its rest energy mpc2, 2then M approaches (2mpK)1/2/c. Therefore, if the energy of the incoming proton is increased by a factor of 9, the mass you can create increases only by a factor of 3, not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so, in principle, all the initial kinetic energy can be used for particle creation. Show that Mc2=2mc2(1+Kmc2) where K is the kinetic energy of each of the two identical colliding particles. Here, if k mc2, we have M directly proportional to K as we would desire.arrow_forward
- (a) Beta decay is nuclear decay in which an electron is emitted. If the electron is given 0.750 MeV of kinetic energy, what is its velocity? (b) Comment on how the high velocity is consistent with the kinetic energy as it compares to the rest mass energy of the electron.arrow_forwardA Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles such as protons. (a) What is the velocity of a proton accelerated by such a potential? (b) An electron?arrow_forwardAs measured by observers in a reference frame S, a particle having charge q moves with velocity v in a magnetic field B and an electric field E. The resulting force on the particle is then measured to be F = q(E + v × B). Another observer moves along with the charged particle and measures its charge to be q also but measures the electric field to be E′. If both observers are to measure the same force, F, show that E′ = E + v × B.arrow_forward
- (a) Calculate for a proton that has a momentum of 1.00 kgm/s. (b) What is its speed? Such protons form a rare component of cosmic radiation with uncertain origins.arrow_forward(a) At what speed will a proton move in a circular path of the same radius as the electron in the previous exercise? (b) What would the radius of the path be if tlie proton had the same speed as the election? (c) What would the radius be if the proton had tlie same kinetic energy' as die electron? (d) The same momentum?arrow_forwardProtons in an accelerator at the Fermi National Laboratory near Chicago are accelerated to an energy of 400 times their rest energy. (a) What is the speed of these protons? (b) What is their kinetic energy in MeV?arrow_forward
- A positron is an antimatter version of the electron, having exactly the same mass. When a positron and an electron meet, they annihilate, converting all of their mass into energy. (a) Find the energy released, assuming negligible kinetic energy before the annihilation. (b) If this energy is given to a proton in the form of kinetic energy, what is its velocity? (c) If this energy is given to another electron in the form of kinetic energy, what is its velocity?arrow_forwardA small charged particle of mass 1.0 x 10-8 kg is traveling rightward between two plates separated by a distance d = 80 cm, as shown below. The electric field between the plates has a constant magnitude of 3.0 x 106 V/m and is directed leftward. The particle's speed is 5.0 x 103 m/s at the left plate and 2.0 x 10³ m/s at the right plate. Ignore the effect of gravity. F (a) Is the particle positively charged or negatively charged? Justify your answer briefly but clearly. (b) Find the charge (with correct sign) of the particle, as well as the potential difference (with correct sign) through which the particle has moved. (Note: The potential difference is positive if the right plate is at a higher potential than the left plate, and negative if the right plate is at a lower potential than the left plate. Show all your work; do not simply plug numbers into a result derived in class.)arrow_forwardA uniform, 2.00 × 106 -V/m electric field is used to accelerate an electron. (a) Calculate the energy the electron acquires once it has moved 40.0 cm across this field. (b) Find the distance it must move across the field to increase its energy by 50.0 GeV.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Electric Fields: Crash Course Physics #26; Author: CrashCourse;https://www.youtube.com/watch?v=mdulzEfQXDE;License: Standard YouTube License, CC-BY