Fisica 4

Student Solutions Manual
for

Modern Physics
Third Edition

Raymond A. Serway Professor Emeritus, James Madison University Clement J. Moses Professor Emeritus, Utica College of Syracuse University Curt A. Moyer University ofNorth Carolina-Wilmington

T H O M S O N

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Contents
Chapter 1: Relativity I Chapter 2: Relativity II Chapter 3: The Quantum Theory of Light Chapter 4: The Particle Nature ofMatter Chapter 5: Matter Waves Chapter 6: Quantum Mechanics in One Dimension Chapter 7: Tunneling Phenomena Chapter 8: Quantum Mechanics in Three Dimensions Chapter 9: Atomic Structure Chapter 10: Statistical Physics Chapter 11: Molecular Structure Chapter 12: The Solid State Chapter 13: Nuclear Structure Chapter 14: Nuclear Physics Applications Chapter 15: Particle Physics

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1Relativity I
1-1 F =—. Consider the special case of constant mass. Then, this equation reduces to FA = maA at in the stationary reference system, and v B = v A + vBA where the subscript A indicates that the measurement is made in the laboratory frame, B the moving frame, and vBA is the velocity of B with respect to A. It is given that a : = —-^-. Therefore from differentiating the velocity atequation, we have aB = a A + aj. Assuming mass is invariant, and the forces are invariant as well, the Newton's law in frame B should be J]F = maA = maB -mav which is not simply ma B . So Newton's second law £ F = W 3 B is invalid in frame B. However, we can rewrite it as X F+maj =ma B , which compares to £ F+mg = m&B. It is as if there were a universal gravitational field g acting on everything. This isthe basic idea of the equivalence principle (General Relativity) where an accelerated reference frame is equivalent to a reference frame with a universal gravitation field. 1-3 IN THE REST FRAME: In an elastic collision energy and momentum are conserved. jji=m1vli+m2v2i=(03kg)(5 Pt=m1vli+m2v2i m/s) + (0.2kg)(-3 m/s) = 0.9 kg m/s

This equation has two unknowns, therefore, apply the conservationof kinetic energy Ei =E{ =—WjUji +—m2v2i =—m1Vif H—m2v2f and conservation of momentum one finds that
Zi £i Jl /U

u lf = -1.31 m/s and v^ = 6.47 m/s or vlf = -1.56 m/s and v^ = 6.38 m/s. The difference in values is due to the rounding off errors in the numerical calculations of the mathematical quantities. If these two values are averaged the values are v1{ = -1.4 m/s and o a = 6.6 m/s, p f...