Physics for Scientists and Engineers with Modern Physics
4th Edition
ISBN: 9780131495081
Author: Douglas C. Giancoli
Publisher: Addison-Wesley
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Chapter 37, Problem 81GP
To determine
The closeness of
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In 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.
From far away a proton is fired directly toward the center of the nucleus of a
mercury atom. Mercury is element number 80, and the diameter of the nucleus is
14.0 fm. Assume the mercury nucleus is fully-ionized, with no electrons bound to it.
If the proton is fired at a speed of 3.8×107 m/s, what is its
closest approach to the surface of the nucleus? Assume the
nucleus remains at rest.
Express your answer with the appropriate units.
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The first line in the Lyman series of H is at 82257.098 cm-1, while that line in the spectrum of deuterium (D) is at 82281.476 cm-1.Check the option that contains the mass of the D nucleus, calculated from this spectral information.
Assume that h = 6,63 x 10-34 J.s, c = 2,99 x 108 m/s, m(elétron) = 9,11 x 10-31 kg e m(próton) = 1,67 x 10-27 kg
Chapter 37 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 37.2 - Prob. 1AECh. 37.2 - Prob. 1BECh. 37.4 - Prob. 1CECh. 37.7 - Prob. 1DECh. 37.7 - Prob. 1EECh. 37.11 - Prob. 1FECh. 37 - Prob. 1QCh. 37 - Prob. 2QCh. 37 - Prob. 3QCh. 37 - Prob. 4Q
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- Explain how a hydrogen atom in the ground state (l = 0) can interact magnetically with an external magnetic field.arrow_forwardAn 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.arrow_forward5arrow_forward
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