Ch41Word - Chapter 41 p. 1 CHAPTER 41 Molecules and Solids...

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Chapter 41 p. 1 CHAPTER 41 – Molecules and Solids Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc / λ = (6.63 × 10 34 J · s)(3.00 × 10 8 m/s)(10 9 nm/m)/(1.60 × 10 19 J/eV) ; E = (1.24 × 10 3 eV · nm)/ . A factor that appears in the analysis of electron energies is e 2 /4p Å 0 = (1.60 × 10 19 C) 2 /4p(8.85 × 10 –12 C 2 /N · m 2 ) = 2.30 × 10 28 J · m. 1. With the reference level at infinity, the binding energy of the two ions is Binding energy = – U = e 2 /4p Å 0 r = ( 2 . 3 0 × 10 28 J · m)/(0.28 × 10 –9 m) = 8.21 × 10 –19 J = 5.1 eV . 2. We convert the units: 1 kcal/mol = (1 kcal/mol)(4186 J/kcal)/(6.02 × 10 23 molecules/mol)(1.60 × 10 19 J/eV) = 0.0435 eV/molecule . For KCl we have (4.43 eV/molecule)[(1 kcal/mol)/(0.0435 eV/molecule)] = 102 kcal/mol . 3. With the repulsion of the electron clouds, the binding energy is Binding energy = – U U clouds ; 4.43 eV = 5.1 eV – U clouds , which gives U clouds = 0.7 eV . 4. When the electrons are midway between the protons, each electron will have a potential energy due to the two protons: U ep = – (2)(0.33) e 2 /4p Å 0 ( r /2) = – (4)(0.33)(2.30 × 10 28 J · m)/(0.074 × 10 9 m)(1.60 × 10 19 J/eV) = – 25.6 eV. The protons have a potential energy: U pp = + e 2 /4p Å 0 r = + (2.30 × 10 28 J · m)/(0.074 × 10 9 m)(1.60 × 10 19 J/eV) = + 19.4 eV. When the bond breaks, each hydrogen atom will be in the ground state with an energy E 1 = – 13.6 eV. Thus the binding energy is Binding energy = 2 E 1 – (2 U ep + U pp ) = 2(–13.6 eV) – [2(– 25.6 eV) + 19.4 eV] = 4.6 eV . 5. The neutral He atom has two electrons in the ground state, n = 1, ¬ = 0, m ¬ = 0. Thus the two electrons have opposite spins, m s = ± ! . If we try to form a covalent bond for the two atoms, we see that an electron from one of the atoms will have the same quantum numbers as one of the electrons on the other atom. From the exclusion principle, this is not allowed, so the electrons cannot be shared. We consider the He 2 + molecular ion to be formed from a neutral He atom and an He + ion. If the electron on the ion has a certain spin value, it is possible for one of the electrons on the atom to have the opposite spin. Thus the electron can be in the same spatial region as the electron on the ion, so a bond can be formed. 6. The units of ˙ 2 / I are (J · s) 2 /(kg · m 2 ) = J 2 /(kg · m/s 2 )m = J 2 /(N · m) = J 2 /J = J.
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Chapter 41 p. 2 7. The reduced mass of the molecule is µ = m 1 m 2 /( m 1 + m 2 ). ( a ) Using data from the periodic table, for NaCl we have = m Na m Cl /( m Na + m Cl ) = (22.9898 u)(35.4527 u)/(22.9898 u + 35.4527 u) = 13.941 u . ( b ) For N 2 we have = m N m N /( m N + m N ) = (14.0067 u)(14.0067 u)/(14.0067 u + 14.0067 u) = 7.0034 u . ( c ) For HCl we have = m H m Cl /( m H + m Cl ) = (1.00794 u)(35.4527 u)/(1.00794 u + 35.4527 u) = 0.9801 u . 8. The reduced mass of the N 2 molecule is N = m N m N /( m N + m N ) = (14.0067 u)(14.0067 u)/(14.0067 u + 14.0067 u) = 7.0034 u. We find the bond length from ˙ 2 /2 I = ˙ 2 /2 N r 2 (2.48 × 10 –4 eV)(1.60 × 10 –19 J/eV) = (1.055 × 10 –34 J · s) 2 /2(7.0034 u)(1.66 × 10 –27 kg/u) r 2 , which gives r = 1.10 × 10 –10 m .
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch41Word - Chapter 41 p. 1 CHAPTER 41 Molecules and Solids...

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