Chapter 31: Atomic PhysicsAnswers to Even-Numbered Conceptual Questions2. There are many such reasons, but perhaps the most important is that orbiting electrons in Rutherford’smodel would radiate energy in the form of electromagnetic waves, with the result that atoms would collapsein a very small amount of time.4. The observation that alpha particles are sometimes reversed in direction when they strike a thin gold foil ledto the idea that there must be a great concentration of positive charge and mass within an atom. This becamethe nucleus in Rutherford’s model.6. In principle, there are an infinite number of spectral lines in any given series. The lines become more closelyspaced as one moves higher in the series, which makes them hard to distinguish in practice.8. (a) There is no upper limit to the wavelength of lines in the spectrum of hydrogen. The reason is that thewavelength is inversely proportional to the energy difference between successive energy levels. The spacingbetween these levels goes to zero as one moves to higher levels, and therefore the correspondingwavelengths go to infinity. (b) There is a lower limit to the wavelength, however, because there is an upperlimit of 13.6 eV to the energy difference between any two energy levels.10. All of these questions can be answered by referring to Figure 31-17 and Table 31-3. (a) Not allowed; there isno d subshell in the n = 2 shell. (b) Not allowed for two reasons. First, there is no p subshell in the n = 1 shell.Second, a p subshell cannot hold seven electrons. (c) Allowed. (d) Not allowed; the n = 4 shell does not have ag subshell.12. No. Atoms in their ground states can emit no radiation. Even if an electron dropped from a highly excitedstate to the ground state in one of these atoms, the result would not be an X-ray. The reason is that thebinding energy of these atoms is simply much lower than the energy of a typical X-ray photon.Solutions to Problems and Conceptual Exercises1. Picture the Problem: In this problem we are given the radius of the nucleus and radius of the atom and want to calculate the fraction of the volume occupied by the nucleus.Strategy: To find the fraction of the volume occupied by the nucleus, divide the volume of the nucleus by the volume ofthe atom. Assume the nucleus and atoms are both spheres.Solution: 1. Write the ratio of volumes in terms of the radius: VnucleusVatom=43πrn343πra3=rnra⎛⎝⎜⎞⎠⎟32. Insert the given radii: VnucleusVatom=0.50×10−15 m5.3×10−11 m⎛⎝⎜⎞⎠⎟3= 8.4×10−16Insight: Most of the mass of the atom is contained in the nucleus, but it accounts for a tiny portion of the total volume.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.31 – 1Chapter 31: Atomic Physics James S. Walker, Physics, 4th Edition2. Picture the Problem: A model of an atom is constructed by using a baseball to represent the nucleus.Strategy: In the previous problem we were told that the diameter of a hydrogen nucleus is 1.0 ×10−15 m and that the electron is typically found about 5.3×10−11 m from the nucleus. Use these values to calculate the ratio of the diameter of the nucleus to the radius of the atom, and then use the ratio to find the model electron’s distance from the center of the baseball.Solution: 1. Set the ratio of the nucleus to atom equal to the ratio of the baseball to the new electrondistance and solve for the electron distance: dnucleusratom=dbaseballrelectron ⇒ relectron=dbaseballdnucleus⎛⎝⎜⎞⎠⎟ratom2. Insert the given distances: relectron=7.3×10–2 m1.00×10–15 m⎛⎝⎜⎞⎠⎟5.3×10−11 m( )= 3.9 kmInsight: If the nucleus were the size of a baseball, the atom would be the size of a small town.3. Picture the Problem: An alpha particle of charge 2e is brought from infinity to the surface of a copper nucleus.Strategy: Let the initial configuration correspond to the alpha particle at rest and infinitely far from the nucleus, and let the final configuration correspond to the alpha particle at rest at a distance of one nuclear radius. The work required to bring the alpha particle near the nucleus is the nonconservative work Wnc=Ef−Ei (equation 8-9), but because the kinetic energy is zero in both configurations, Wnc=ΔU.Use equation 20-8 to solve for the change in potential energy.Solution: 1. (a) Find the work required to bring an alpha particle from infinity to a distance r: Wnc=ΔU =kq0qr−0⎛⎝⎜⎞⎠⎟=k 2e( )29e( )r=58ke2r2. Insert the constants and set the distance equal to half the diameter of the nucleus: W =58 8.99×109 N ⋅m2C2( )1.60×10–19 C( )2124.8×10–15 m( )= 5.6 pJInsight: The energy may seem small, but it is equivalent to 35 MeV, over 68 times the rest energy of an electron!4. Picture the Problem: The image shows an alpha particle approaching a gold nucleus with an initial kinetic energy of 3.0 MeV. The alpha particle comes to rest at a distance d from the gold nucleus when all of the kinetic energy has been converted to electric potential energy. Strategy: Set the initial kinetic energy equal to the potential energy at the point d, using equation 20-8. Solve the resulting equation for the distance d.Solution: 1. Set the initial kinetic energy equal to the final potential energy: K =kqQd2. Solve for the separation distance: d =kqQK=k(2e)(79e)K=158ke2K3. Insert the given data and constants: d =158 8.99×109N ⋅m2C2( )1.60×10–19C( )23.0×106eV( )1.60×10–19 J/eV( )= 76 fmInsight: Because the alpha particle moves away from the nucleus without colliding, this experiment verifies that the gold nucleus has a radius smaller than 76 fm.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.31 – 2Chapter 31: Atomic Physics James S. Walker, Physics, 4th EditionCopyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.31 –
View Full Document
Unlocking...