Chapter 32: Nuclear Physics and Nuclear RadiationAnswers to Even-Numbered Conceptual Questions2. The difference is that in an decay only a single particle is emitted—the particle—and it carries the energyreleased by the decay. In the case of decay, two particles are emitted—the particle (electron) and thecorresponding antineutrino. These two particles can share the energy of decay in different amounts, whichaccounts for the range of observed energies for the particles. (Of course, the antineutrinos are very difficultto detect.)4. Alpha particles, which can barely penetrate a sheet of paper, are very unlikely to expose film in a cardboardbox. Beta particles, on the other hand, are able to penetrate a few millimeters of aluminum. Therefore, betaparticles are more likely to expose the film than alpha particles.6. A change in isotope is simply a change in the number of neutrons in a nucleus. The electrons in the atom,however, respond only to the protons with their positive charge. Because electrons are responsible forchemical reactions, it follows that chemical properties are generally unaffected by a change in isotope.8. Above the N = Z line, a nucleus contains more neutrons than protons. This helps to make the nucleus stable,by spreading out the positive charge of the protons. If a nucleus were below the N = Z line, it would havemore protons than neutrons, and electrostatic repulsion would blow the nucleus apart.10. No. Fossil dinosaur skeletons represent organic material—which is necessary for carbon-14 dating—but theyare thousands of times too old for the technique to be practical.12. Yes. If the different isotopes have different decay rates—which is generally the case—they can still have thesame activity if they are present in different amounts.Solutions to Problems and Conceptual Exercises1. Picture the Problem: We are given three isotope designations and asked to calculate the atomic number (A), neutron number (N), and atomic mass (Z) for each isotope.Strategy: Use the designation of an isotope is ZAX , where A is the atomic mass, Z is the atomic number, and X is the chemical designation, to determine the atomic number and atomic mass of the three isotopes. Calculate the neutron number by subtracting the atomic number from the atomic mass. Solution: 1. (a) Write Z, N, and A for the isotope 92238U : Z = 92 N =A−Z =238−92= 146 A = 2382. (b) Write Z, N, and A for the isotope 94239Pu: Z = 94 N =A−Z =239−94= 145 A = 2393. (c) Write Z, N, and A for the isotope 60144Nd: Z = 60 N =A−Z =144−60= 84 A = 144Insight: Note that for each of these isotopes the neutron number is about 1½ times the atomic number. The neutron number is about equal to the atomic number for small isotopes and about 1½ times the atomic number for large isotopes.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.32 – 1Chapter 32: Nuclear Physics and Nuclear Radiation 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.32 – 2Chapter 32: Nuclear Physics and Nuclear Radiation James S. Walker, Physics, 4th Edition2. Picture the Problem: We are given three isotope designations and asked to calculate the atomic number (A), neutron number (N), and atomic mass (Z) for each isotope.Strategy: Use the designation of an isotope is ZAX , where A is the atomic mass, Z is the atomic number, and X is the chemical designation, to determine the atomic number and atomic mass of the three isotopes. Calculate the neutron number by subtracting the atomic number from the atomic mass. Solution: 1. (a) Write Z, N, and A for the isotope 80202Hg: Z = 80 N =A−Z =202−80= 122 A = 2022. (b) Write Z, N, and A for the isotope 86220Rn: Z = 86 N =A−Z =220−86= 134 A = 2203. (c) Write Z, N, and A for the isotope 4193Nb: Z = 41 N =A−Z =93−41= 52 A = 93Insight: Note that for each of these isotopes, the neutron number is about 1½ times the atomic number. The neutron number is about equal to the atomic number for small isotopes and about 1½ times the atomic number for large isotopes.3. Picture the Problem: The nuclear radius increases as the atomic mass number increases.Strategy: Insert the mass number (A) from each isotope into equation 32-4 to calculate the nuclear radius. Solution: 1. (a) Calculate the radius of 79197Au(A = 197): r = 1.2×10−15 m( )A1/3= 1.2 fm( )197( )1/3= 7.0 fm2. (b) Calculate the radius of 2760Co(A = 60): r = 1.2 fm( )60( )1/3= 4.7 fmInsight: 79197Auhas over three times the number of nucleons as 2760Co, yet its radius in not even twice as large.4. Picture the Problem: The number of neutrons in a chlorine nucleus can be determined from its nuclear radius.Strategy: Use equation 32-4 to calculate the mass number of the chlorine atom. The periodic table shows that the atomic number of chlorine is 17. Subtract the atomic number from the atomic mass to calculate the neutron number.Solution: 1. Solve equation 32-4 for the atomic mass: A =r31.2×10−15 m( )3=4.0×10−15 m1.2×10−15 m⎛⎝⎜⎞⎠⎟3=372. Solve equation 32-1 for the neutron number: A =Z +N ⇒ N =A−Z =37−17= 20Insight: At best we can say there are between 19 and 21 neutrons because the radius is only given to two significant figures. For instance, note that adding one additional neutron to make A = 38 also gives a radius of 4.0 ×10−15 m.5. Picture the Problem: The nuclear densities of thorium-238 and an alpha particle (helium-4) can be compared.Strategy: The nuclear density is the mass of the nucleus divided by its volume. Calculate the mass by multiplying the mass number by the mass of one nucleon. The volume is the volume of a sphere whose radius is given by equation 33-4.Solution: 1. (a) Write the density as mass over volume: ρ =MV=Am43πr3Copyright © 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
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