An alpha particle with kinetic energy 11.0 MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=pob, where po is the magnitude of the initial momentum of the alpha particle and b=1.50x10-12 m. (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.) Repeat for b=1.10x10-13 m Express your answer in meters. ΑΣφ m Submit Request Answer Part C Repeat for b=1.50x10-14 m. Express your answer in meters. m
An alpha particle with kinetic energy 11.0 MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=pob, where po is the magnitude of the initial momentum of the alpha particle and b=1.50x10-12 m. (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.) Repeat for b=1.10x10-13 m Express your answer in meters. ΑΣφ m Submit Request Answer Part C Repeat for b=1.50x10-14 m. Express your answer in meters. m
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:An alpha particle with kinetic energy 11.0 Me V
makes a collision with lead nucleus, but it is not
"aimed" at the center of the lead nucleus, and has
an initial nonzero angular momentum (with respect
to the stationary lead nucleus) of magnitude
L%=pob, where po is the magnitude of the initial
momentum of the alpha particle and
b=1.50x10-12m (Assume that the lead nucleus
remains stationary and that it may be treated as a
point charge. The atomic number of lead is 82. The
alpha particle is a helium nucleus, with atomic
number 2.)
Repeat for b=1. 10×10-13 m.
Express your answer in meters.
ΑΣφ
Submit
Request Answer
Part C
Repeat for b=1.50×10-14 m.
Express your answer in meters.
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