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CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
Fundamental Concepts Electrons in Atoms 2.1 Cite the difference between atomic mass and atomic weight.Solution Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally occurring isotopes.(Materials Science and Engineering An Introduction, Enhanced eText, 10e William Callister, David Rethwisch) (For Complete File, Download link at the end of this File) NOTE: There is no Solution Manual for Chapter 1 1 / 4
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2.2 Chromium has four naturally-occurring isotopes: 4.34% of
50 Cr, with an atomic weight of 49.9460 amu, 83.79% of 52 Cr, with an atomic weight of 51.9405 amu, 9.50% of 53 Cr, with an atomic weight of 52.9407 amu, and 2.37% of 54 Cr, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic weight of Cr is 51.9963 amu.
Solution The average atomic weight of chromium is computed by adding fraction-of-occurrence/atomic weight products for the three isotopes. Thus
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2.3 Hafnium has six naturally occurring isotopes: 0.16% of
174 Hf, with an atomic weight of 173.940 amu; 5.26% of 176 Hf, with an atomic weight of 175.941 amu; 18.60% of 177 Hf, with an atomic weight of 176.943 amu; 27.28% of 178 Hf, with an atomic weight of 177.944 amu; 13.62% of 179 Hf, with an atomic weight of 178.946 amu;.and 35.08% of 180 Hf, with an atomic weight of 179.947 amu. Calculate the average atomic weight of Hf.
Solution The average atomic weight of halfnium is computed by adding fraction-of-occurrence—atomic weight products for the six isotopes—i.e., using Equation 2.2. (Remember: fraction of occurrence is equal to the percent of occurrence divided by 100.) Thus
Including data provided in the problem statement we solve for as
= 178.485 amu
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2.4 Bromium has two naturally occurring isotopes:
79 Br, with an atomic weight of 78.918 amu, and 81 Br, with an atomic weight of 80.916 amu. If the average atomic weight for Br is 79.903 amu, calculate the fraction-of- occurrences of these two isotopes.
Solution The average atomic weight of indium is computed by adding fraction-of-occurrence—atomic weight products for the two isotopes—i.e., using Equation 2.2, or
Because there are just two isotopes, the sum of the fracture-of-occurrences will be 1.000; or
which means that
Substituting into this expression the one noted above for , and incorporating the atomic weight values provided in the problem statement yields
Solving this expression for yields . Furthermore, because
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