An Introduction to Physical Science
14th Edition
ISBN: 9781305079137
Author: James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher: Cengage Learning
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Chapter 9, Problem 11MC
To determine
The pair of particle properties which cannot be determined exactly and simultaneously.
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Do you know what characteristic of the graph corresponds to total energy?
(d) Identify which two particles have the same total energy.
3.
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Ernest Rutherford is famous for, among other things, shooting alpha particles at unsuspecting gold atoms. Consider an alpha particle endowed with 5.00 MeV of energy. Determine the closest distance this particle can approach the nucleus of a gold atom before deflecting
where 1 eV = 1.602 × 10-19 J. Express the neutron’s kinetic energy in electron volts.
b) In nuclear physics, it is convenient to express the energy of particles in electron volts (eV),
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In the tendon at this pon
a) Find the kinetic energy of the neutron in joules
Chapter 9 Solutions
An Introduction to Physical Science
Ch. 9.1 - Prob. 1PQCh. 9.1 - Prob. 2PQCh. 9.2 - Prob. 1PQCh. 9.2 - Prob. 2PQCh. 9.2 - Prob. 9.1CECh. 9.3 - Prob. 1PQCh. 9.3 - When does a hydrogen atom emit or absorb radiant...Ch. 9.3 - Prob. 9.2CECh. 9.3 - Prob. 9.3CECh. 9.3 - Prob. 9.4CE
Ch. 9.4 - Prob. 1PQCh. 9.4 - Prob. 2PQCh. 9.5 - Prob. 1PQCh. 9.5 - Prob. 2PQCh. 9.6 - Prob. 1PQCh. 9.6 - Prob. 2PQCh. 9.6 - Prob. 9.5CECh. 9.7 - Prob. 1PQCh. 9.7 - Prob. 2PQCh. 9 - Prob. AMCh. 9 - Prob. BMCh. 9 - Prob. CMCh. 9 - Prob. DMCh. 9 - Prob. EMCh. 9 - Prob. FMCh. 9 - Prob. GMCh. 9 - Prob. HMCh. 9 - Prob. IMCh. 9 - Prob. JMCh. 9 - Prob. KMCh. 9 - Prob. LMCh. 9 - Prob. MMCh. 9 - Prob. NMCh. 9 - Prob. OMCh. 9 - Prob. PMCh. 9 - Prob. QMCh. 9 - Prob. 1MCCh. 9 - Prob. 2MCCh. 9 - Prob. 3MCCh. 9 - Prob. 4MCCh. 9 - Prob. 5MCCh. 9 - Prob. 6MCCh. 9 - Prob. 7MCCh. 9 - Prob. 8MCCh. 9 - Prob. 9MCCh. 9 - Prob. 10MCCh. 9 - Prob. 11MCCh. 9 - Prob. 12MCCh. 9 - Prob. 13MCCh. 9 - Prob. 14MCCh. 9 - Prob. 1FIBCh. 9 - Prob. 2FIBCh. 9 - Prob. 3FIBCh. 9 - Prob. 4FIBCh. 9 - Prob. 5FIBCh. 9 - Prob. 6FIBCh. 9 - Prob. 7FIBCh. 9 - Prob. 8FIBCh. 9 - Prob. 9FIBCh. 9 - Prob. 10FIBCh. 9 - Prob. 11FIBCh. 9 - Prob. 12FIBCh. 9 - Prob. 1SACh. 9 - Prob. 2SACh. 9 - Prob. 3SACh. 9 - Prob. 4SACh. 9 - Prob. 5SACh. 9 - Prob. 6SACh. 9 - Prob. 7SACh. 9 - Prob. 8SACh. 9 - Prob. 9SACh. 9 - Prob. 10SACh. 9 - Prob. 11SACh. 9 - Prob. 12SACh. 9 - Prob. 13SACh. 9 - Prob. 14SACh. 9 - Prob. 15SACh. 9 - Prob. 16SACh. 9 - Prob. 17SACh. 9 - Prob. 18SACh. 9 - Prob. 19SACh. 9 - Prob. 20SACh. 9 - Prob. 21SACh. 9 - Prob. 22SACh. 9 - Prob. 23SACh. 9 - Prob. 24SACh. 9 - Prob. 25SACh. 9 - Prob. 26SACh. 9 - Prob. 27SACh. 9 - Prob. 28SACh. 9 - Prob. 29SACh. 9 - Prob. 30SACh. 9 - Prob. 31SACh. 9 - Prob. 32SACh. 9 - Prob. 33SACh. 9 - Prob. 34SACh. 9 - Visualize the connection for the descriptions of...Ch. 9 - Prob. 1AYKCh. 9 - Prob. 2AYKCh. 9 - Prob. 3AYKCh. 9 - Prob. 4AYKCh. 9 - Prob. 5AYKCh. 9 - Prob. 1ECh. 9 - Prob. 2ECh. 9 - Prob. 3ECh. 9 - Prob. 4ECh. 9 - Prob. 5ECh. 9 - Prob. 6ECh. 9 - Prob. 7ECh. 9 - Prob. 8ECh. 9 - Prob. 9ECh. 9 - Prob. 10ECh. 9 - Prob. 11ECh. 9 - Prob. 12E
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- A proton moving in the positive x direction at 4.3 Mm/s collides with a nucleus. The collision lasts 0.12 fs, and the average impulsive force is 42 i + 17 j micro - Newton. A) Find the velocity of the proton after the collision. B) Through what angle has the proton's motion been deflected?arrow_forwardConsidering electron and proton as two charged particles separated by d = 5.9 × 10-11 m calculate the gravitational force between the proton and electron and find its ratio to the Coulomb force. Take the mass of the proton 1.7 x 10-27 kg, the mass of the electron 9.1 x 10-31 kg, the value of = 9x10⁹ m/F. Give the answer for the universal gravitational constant 6.7 x 10-11 N kg 2m-2, the electron charge -1.6 x 10-¹9 C and the gravitational force in 10-47 N. 1 Απερ Answer:arrow_forward#Q33 I need quick answer.arrow_forward
- An electron accelerated in an x-ray tube hits an anode (positively charged plate) target. V= HOTE m/s +High V- = anode kV cer cer Ger 4848 46 N Now 484846 a) Assuming all of its energy is transferred in generating x-rays, how fast is the electron moving when it reaches the anode target if it produces an x-ray with an energy of 46 keV . The mass of the electron is 9.1 x10-31 kg. IM M Hint: How is the energy of x-rays related to the KE of the electrons? How is the kinetic energy related to the speed? X-rays cathode b) What is the accelerating potential of the x-ray tube (potental difference between the anode and cathode) in part a)? Va Ponder: How large is this volatge? Remember the a regular AA battery has a voltage of 1.5 V.arrow_forwardA linear particle accelerator using beta particles collides electrons with their anti-matter counterparts, positrons. The accelerated electron hits the stationary positron with a velocity of 19 x 106 m/s, causing the two particles to annihilate.If two gamma photons are created as a result, calculate the energy of each of these two photons, giving your answer in MeV (mega electron volts), accurate to 1 decimal place. Take the mass of the electron to be 5.486 x 10-4 u, or 9.109 x 10-31 kg.Note: Assume that the kinetic energy is also converted into the gamma rays, and is included in the two photons.arrow_forwardThe very high speeds of alpha particles make them suitable for experiments that probe the nature of matter. A nucleus ejects an alpha particle with a kinetic energy of 8.3 MeV, a typical energy. How fast is the alpha particle moving?arrow_forward
- Options: a) Neither, both have the same final KE b) the proton The answer is (a) neither, but I am struggling to understand why. I included my work, so maybe you can help me understand where I am going wrong. Things we know: KE = 1/3 MV^2 Mass of electron : 9.1 x 10^-31 kg (smaller) Mass of proton: 1.67 x 10^-27 kg (larger) Force = same magnitude for both since they both have the same magnitude of charge in the same Efield F=maarrow_forwarda) Add the missing particles required to satisfy our laws of physics. p = uud, n = ddu, л¹ = ud, π = ud, лº = uu or dd. T → e + n→ p+ T° → e+v p+n→p+p+p+ b) For reactions (1) and (4) in part a), can the initial particles be at rest? Why/why not?arrow_forwardNote: m=93 4. A particle of charge q = moves in velocity i = (2+ m)î - (5 + m)j – 4k in magnetic field B = (7+ m)j + 2k and electric field Ē = 5î – j + (2 + m)k. a. Compute the Lorentz force experienced by the particle b. The magnitude of this forcearrow_forward
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