modernphysics - Chapter 10 The Problem Solutions 1. The ion...

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Inha University Department of Physics Chapter 10 The Problem Solutions 1. The ion spacings and melting points of the sodium halides are as follows: 660 740 801 988 Melting point, o C 0.32 0.29 0.28 0.23 Ion spacing, nm Nal NaBr NaCl NaF Explain the regular variation in these quantities with halogen atomic number. Sol The halogenic atomic numbers are Z = 9 for fluorine (F), Z = 17 for chlorine (Cl), Z =35 for bromine (Br) and Z = 53 for iodine (I). The greater the atomic number of a halogen atom, the larger the atom is, hence the increase in the interatomic spacing with Z . The larger the ion spacing, the smaller the cohesive energy, hence the lower the melting point. 3. (a) The ionization energy of potassium is 4.34 eV and the electron affinity of chlorine is 3.61 eV. The Madelung constant for the KCl structure is 1.748 and the distance between ions of opposite sign is 0.314 nm. On the basis of these data only, compute the cohesive energy of KCl. (b) The observed cohesive energy of KCl is 6.42 eV per ion pair. On the assumption that the difference between this figure and that obtained in a is due to the exclusion-principle repulsion, find the exponent n in the formula Br -n for the potential energy arising from this source.
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Inha University Department of Physics Sol (a) The cohesive energy will be the negative of the Coulombic energy as given in Equation (10.1), minus the difference between the ionization energy of potassium and the electron affinity of chlorine; ( 29 eV 29 7 eV 61 3 eV 34 4 J/eV 10 602 1 m 10 314 0 C 10 602 1 /C m N 10 988 8 748 1 4 19 9 2 19 2 2 9 2 . ) . . ( ) . )( . ( ) . )( . ( ) . ( = - - × × × × = - - - - - a i o E E r e pe a (b) The difference between the observed binding energy and that found in part (a) must be due to the repulsive energy as given in Equation (10. 1). From the observed binding energy, U O must be given by - U O = 6.42 eV + (4.34 eV - 3.61 eV) = 7.15 eV. The Couloumbic energy, an intermediate calculation in part (a) , is U coulomb -8.0156 eV, and so solving Equation (10.5) for n , 26 9 eV 0156 8 eV 15 7 1 1 1 1 coulomb . . . = - = - = - - U U n o
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Inha University Department of Physics Sol The heat lost by the expanding gas is equal to the work done against the attractive van der Waals forces between the gas molecules. 5.
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modernphysics - Chapter 10 The Problem Solutions 1. The ion...

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