# Ch40Word - Chapter 40 p 1 CHAPTER 40 Quantum Mechanics of...

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Chapter 40 p. 1 CHAPTER 40 – Quantum Mechanics of Atoms 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 × 10 8 m/s)(10 9 nm/m)/(1.60 × 10 –19 J/eV) ; E = (1.24 × 10 3 eV · nm)/ . 1. The value of ¬ can range from 0 to n – 1. Thus for n = 6, we have ¬ = 0, 1, 2, 3, 4, 5 . 2. The value of m ¬ can range from – ¬ to + ¬ . Thus for ¬ = 3, we have m ¬ = – 3, – 2, – 1, 0, 1, 2, 3 . The possible values of m s are ! , + ! . 3. The value of ¬ can range from 0 to n – 1. Thus for n = 4, we have ¬ = 0, 1, 2, 3. For each ¬ the value of m ¬ can range from – ¬ to + ¬ , or 2 ¬ + 1 values. For each of these there are two values of m s . Thus the total number for each ¬ is 2(2 ¬ + 1). The number of states is N = 2(0 + 1) + 2(2 + 1) + 2(4 + 1) + 2(6 + 1) = 32 states . We start with ¬ = 0, and list the quantum numbers in the order ( n , ¬ , m ¬ , m s ); (4, 0, 0, – ! ), (4, 0, 0, + ! ), (4, 1, –1, – ! ), (4, 1, –1, + ! ), (4, 1, 0, – ! ), (4, 1, 0, + ! ), (4, 1, 1, – ! ), (4, 1, 1, + ! ), (4, 2, – 2, – ! ), (4, 2, – 2, + ! ), (4, 2, –1, – ! ), (4, 2, –1, + ! ), (4, 2, 0, – ! ), (4, 2, 0, + ! ), (4, 2, 1, – ! ), (4, 2, 1, + ! ), (4, 2, 2, – ! ), (4, 2, 2, + ! ), (4, 3, – 3, – ! ), (4, 3, – 3, + ! ), (4, 3, – 2, – ! ), (4, 3, – 2, + ! ), (4, 3, – 1, – ! ), (4, 3, – 1, + ! ), (4, 3, 0, – ! ), (4, 3, 0, + ! ), (4, 3, 1, – ! ), (4, 3, 1, + ! ), (4, 3, 2, – ! ), (4, 3, 2, + ! ), (4, 3, 3, – ! ), (4, 3, 3, + ! ). 4. The value of m ¬ can range from – ¬ to + ¬ , so we have ¬ = 3 . The value of ¬ can range from 0 to n – 1. Thus we have n = ¬ + 1 (minimum 4). There are two values of m s : m s = – ! , + ! . 5. The value of ¬ can range from 0 to n – 1. Thus for ¬ = 4, we have n = 5 . For each ¬ the value of m ¬ can range from – ¬ to + ¬ : m ¬ = – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4 . There are two values of m s : m s = – ! , + ! . 6. The magnitude of the angular momentum depends on ¬ only: L = ˙ [ ¬ ( ¬ + 1)] 1/2 = (1.055 × 10 –34 J · s)[(2)(2 + 1)] 1/2 = v6 ˙ = 2.58 × 10 –34 kg · m 2 /s . 7. ( a ) The principal quantum number is n = 6 . ( b ) The energy of the state is E 6 = – (13.6 eV)/ n 2 = – (13.6 eV)/6 2 = – 0.378 eV . ( c ) From spdfgh, we see that the “g” subshell has ¬ = 4 . The magnitude of the angular momentum d e p e n d s o n ¬ only: L = ˙ [ ¬ ( ¬ + 1)] 1/2 = (1.055 × 10 –34 J · s)[(4)(4 + 1)] 1/2 = v20 ˙ = 4.72 × 10 –34 kg · m 2 /s . ( d ) For each ¬ the value of m ¬ can range from – ¬ to + ¬ : m ¬ = – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4 .

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Chapter 40 p. 2 8. ( a ) For each ¬ the value of m ¬ can range from – ¬ to + ¬ , or 2 ¬ + 1 values. For each of these there are two values of m s . Thus the total number of electrons allowed in a subshell is N = 2(2 ¬ + 1). ( b ) For the values of ¬ we have ¬ = 0, N = 2[2(0) + 1] = 2; ¬ = 1, N = 2[2(1) + 1] = 6; ¬ = 2, N = 2[2(2) + 1] = 10; ¬ = 3, N = 2[2(3) + 1] = 14; ¬ = 4, N = 2[2(4) + 1] = 18; ¬ = 5, N = 2[2(5) + 1] = 22; ¬ = 6, N = 2[2(6) + 1] = 26 . 9. From Problem 8 the total number of electrons allowed in a subshell with quantum number ¬ is 2(2 ¬ + 1). For a state with quantum number n the value of ¬ can range from 0 to n – 1. Thus we have N =2 (2 +1) Σ =0 n – 1 =4 Σ n – 1 +2 1 Σ n – 1 ( n –1) n 2 n n 2 .
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Ch40Word - Chapter 40 p 1 CHAPTER 40 Quantum Mechanics of...

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