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homework7.pdf - Fall 2012 Reynolds PY 203 Modern Physics...

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Fall 2012 PY 203: Modern Physics Reynolds Solutions for Problem Set 9 Due Wednesday, November 7 1. The dissociation energy of Cl 2 is 2.48 eV. Consider the formation of an NaCl molecule by the reaction Na + 1 2 Cl 2 NaCl . Is this reaction endothermic (requiring energy) or exothermic (releasing energy)? How much energy per molecule is required or given off? Solution. Strategy: Use the numbers from Figure 37-1, adding the preliminary step of dissociating the Cl molecule. So: In Figure 37-1, the dissociation energy (net gain starting from neutral Na and Cl atoms and ending with NaCl) is given as 4.26 eV. The additional cost you have to pay is that of liberating a chlorine atom, or 2.48/2 eV (The factor of 2: if you dissociate one Cl 2 molecule, for a price of 2.48 eV, you get to make two NaCl molecules, so the extra price per molecule is 1.24 eV.) So the reaction is still exothermic , but since you had to pay extra at the beginning, you get less out: 4.26 - 1.24 = 3.02 eV.

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2 2. The equilibrium separation of the Rb + and Cl ions in RbCl is about 0.267 nm. (a) Calculate the potential energy of attraction of the ions, assuming them to be point charges. (b) The ionization energy of rubidium is 4.18 eV, and the electron affinity of Cl is 3.62 eV. Find the dissociation energy, neglecting the energy of repulsion. (c) The measured dissociation energy is 4.37 eV. What is the energy due to repulsion of the ions? Solution. Strategy: Use the Coulomb potential for (a), and for the bookkeeping, use the reasoning of Example 37-1. So: (a) Just the Coulomb potential: U = - ke 2 r = - (9 × 10 9 Nm 2 C 2 )(1 . 6 × 10 19 C) 2 . 67 × 10 10 m = - 8 . 6 × 10 19 J = - 5 . 39 eV . (It is legitimate to drop the minus sign and include that information in words, something like, “The potential energy of attraction is 5.39 eV,” because the word “attraction” conveys that the P.E. is negative.) (b) The net cost of making the Rb + and Cl ions is 4.18 – 3.62 = 0.56 eV. This is deducted from the potential energy of attraction to make the dissociation energy (if we neglect repulsion) which is therefore 5.39 – 0.56 = 4.83 eV. (To be finicky, we could write - 5 . 39 + 0 . 56 = - 4 . 83 eV, but conventionally we just quote the magnitude when we talk about dissociation energies. If the energy were really positive, the molecule wouldn’t be bound!) (c) The real dissociation energy includes the effects of (a) and (b) as well as extra positive energy due to the repulsion – that is, it’s less in absolute value than the result of (b). We’re told that it is really 4.37 eV, that is, the repulsive energy that we neglected in (b) must be 4.83 – 4.37 = 0.46 eV.
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• Fall '08
• Reynolds,S
• Atom, Chemical bond, Chemical polarity, dipole moment, Solution. Strategy

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