College Physics (10th Edition)
10th Edition
ISBN: 9780321902788
Author: Hugh D. Young, Philip W. Adams, Raymond Joseph Chastain
Publisher: PEARSON
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Chapter 28, Problem 8CQ
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The reason why the name wavicles is an appropriate term for particles such as electrons.
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Determine the distance between the electron and proton in an atom if the potential energy U of the electron is 13.8 ev (electronvolt, 1 eV
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Spectral Analysis. While studying the spectrum of a gas cloud in space, an astronomer magnifies a spectral line that results from a transition from a p state to an s state. She finds that the line at 575.050 nm has actually split into three lines, with adjacent lines 0.0462 nm apart, indicating that the gas is in an external magnetic field. (Ignore effects due to electron spin.) What is the strength of the external magnetic field?
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Chapter 28 Solutions
College Physics (10th Edition)
Ch. 28 - Prob. 1CQCh. 28 - Prob. 2CQCh. 28 - Prob. 3CQCh. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQCh. 28 - Prob. 8CQCh. 28 - Prob. 9CQCh. 28 - Prob. 10CQ
Ch. 28 - Prob. 11CQCh. 28 - Prob. 12CQCh. 28 - Prob. 1MCPCh. 28 - Prob. 2MCPCh. 28 - Prob. 3MCPCh. 28 - Prob. 4MCPCh. 28 - Prob. 5MCPCh. 28 - Prob. 6MCPCh. 28 - Prob. 7MCPCh. 28 - Prob. 8MCPCh. 28 - Prob. 9MCPCh. 28 - Prob. 10MCPCh. 28 - Prob. 11MCPCh. 28 - Prob. 12MCPCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 5PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Prob. 13PCh. 28 - Prob. 14PCh. 28 - Prob. 15PCh. 28 - Prob. 16PCh. 28 - Prob. 17PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 22PCh. 28 - Prob. 23PCh. 28 - Prob. 24PCh. 28 - Prob. 25PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 28PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 31PCh. 28 - Prob. 32PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 38PCh. 28 - Prob. 39PCh. 28 - Prob. 40PCh. 28 - Prob. 41PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - Prob. 45PCh. 28 - Prob. 46PCh. 28 - Prob. 47PCh. 28 - Prob. 48PCh. 28 - Prob. 49PCh. 28 - Prob. 50GPCh. 28 - Prob. 51GPCh. 28 - Prob. 52GPCh. 28 - Prob. 53GPCh. 28 - Prob. 54GPCh. 28 - Prob. 55GPCh. 28 - Prob. 56GPCh. 28 - Prob. 57GPCh. 28 - Prob. 58GPCh. 28 - Prob. 59GPCh. 28 - Prob. 61GPCh. 28 - Prob. 62GPCh. 28 - Prob. 63GPCh. 28 - Prob. 64GPCh. 28 - Prob. 65GPCh. 28 - Prob. 66PPCh. 28 - Prob. 67PPCh. 28 - Prob. 68PPCh. 28 - Prob. 69PPCh. 28 - Prob. 70PPCh. 28 - Prob. 71PPCh. 28 - Prob. 72PPCh. 28 - Prob. 73PPCh. 28 - Prob. 74PP
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- A Thomson-type experiment with relativistic electrons. One of the earliest experiments to show that p = mv (rather than p = mv) was that of Neumann. [G. Neumann, Ann. Physik 45:529 (1914)]. The apparatus shown in Figure P4.5 is identical to Thomsons except that the source of high-speed electrons is a radioactive radium source and the magnetic field B is arranged to act on the electron over its entire trajectory from source to detector. The combined electric and magnetic fields act as a velocity selector, only passing electrons with speed v, where v = V/Bd (Equation 4.6), while in the region where there is only a magnetic field the electron moves in a circle of radius r, with r given by p = Bre. This latter region (E = 0, B = constant) acts as a momentum selector because electrons with larger momenta have paths with larger radii. (a) Show that the radius of the circle described by the electron is given by r = (l2 + y2)/2y. (b) Typical values for the Neumann experiment were d = 2.51 104 m, B = 0.0177 T, and l = 0.0247 m. For V = 1060 V, y, the most critical value, was measured to be 0.0024 0.0005 m. Show that these values disagree with the y value calculated from p = mv but agree with the y value calculated from p = mv within experimental error. (Hint: Find v from Equation 4.6, use mv = Bre or mv = Bre to find r, and use r to find y.) Figure P4.5 The Neumann apparatus.arrow_forwardWhat is the magnitude of the spin momentum of an electron? (Express you answer in terms of h.)arrow_forwardThe force on an electron is “negative the gradient of the potential energy function.” Use this knowledge and Equation 8.1 to show that the force on the electron in a hydrogen atom is given by Coulomb’s force law. Ur=ke2r(8.1)arrow_forward
- Consider a hydrogen-like ion where an electron is orbiting a nucleus that has charge q=+Ze. Derive the formulas for the energy En of the electron in nth orbit and the orbital radius rn.arrow_forwardA beryllium ion with a single electron (denoted Be3+) is in an excited state with radius the same as that of the ground state of hydrogen. (a) What is n for the Be3+ ion? (b) How much energy in eV is needed to ionize the ion from this excited state?arrow_forwardWhat is the greatest magnitude of the orbital angular momentum L that you can find in a state vith n= 6? O6 h O5.48 h O5.92 h O6.48 harrow_forward
- 8. A white dwarf is a very dense star, with its ions and electrons packed extremely close together. Each electron may be considered to be located within a region of size x= 1.5 x 10-12 m. Estimate the minimum speed of the electron. Do you think that the effects of relativity will be important for these stars?arrow_forwardAn electron revolve round the nucleus with the radius of the circular orbit is 'r'. To double the kinetic energy of the electron its orbital radius will be. v/ /2 (b) V2 r, (a) (c) 2r (d) r/2arrow_forwardIf the nucleus is a few fm in diameter, the distance between the centers of two protons must be ≈2 fm.a. Calculate the repulsive electric force between two protons that are 2.0 fm apart.b. Calculate the attractive gravitational force between two protons that are 2.0 fm apart. Could gravity be the force that holds the nucleus together?arrow_forward
- Determine the distance between the electron and proton in an atom if the potential energy U of the electron is 10.1 eV (electronvolt, 1 eV = 1.6 × 10-19 J). Give your answer in Angstrom (1 A = 10-10 m). Answer: Choose... +arrow_forwardA doubly ionized lithium atom (Li++) is one that has had two of its three electrons removed. The energy levels of the remaining single-electron ion are closely related to those of the hydrogen atom. The nuclear charge for lithium is +3e instead of just +e. How are the energy levels related to those of hydrogen? How is the radius of the ion in the ground level related to that of the hydrogen atom? Explain.arrow_forwardWhat two pieces of evidence allowed the first calculation of me , the mass of the electron?(a) The ratios qe / me and q p / m p .(b) The values of qe and EB .(c) The ratio qe / me and qe .Justify your response.arrow_forward
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