1131_PartUniversity Physics Solution

1131_PartUniversity Physics Solution - 42-12 Chapter 42(ii...

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42-12 Chapter 42 (ii) transition 1 0, 2 1 n n l l = = = = 2 2 2 (6 2) 2 E hf hf I I Δ = + = + = = 8 11 13 2.998 10 m/s 4.28 m 2( / 2 ) 2(3.960 10 Hz) 6.93 10 Hz c I f λ μ π × = = = + × + × = (iii) transition 2 1, 2 3 n n l l = = = = 2 2 3 (6 12) 2 E hf hf I I Δ = + = − + = = 8 11 13 2.998 10 m/s 4.40 m 3( / 2 ) 3(3.960 10 Hz) 6.93 10 Hz c I f λ μ π × = = = + × + × = E VALUATE : The vibrational energy change for the 1 0 n n = = transition is the same as for the 2 1 n n = = transition. The rotational energies are much smaller than the vibrational energies, so the wavelengths for all three transitions don’t differ much. 42.48. The sum of the probabilities is / F F 1 1 1 ( ) ( ) 1. 1 1 1 1 E kT E kT E kT E kT E kT e f E E f E E e e e e −Δ −Δ Δ −Δ −Δ + Δ + − Δ = + = + = + + + + 42.49. Since potassium is a metal we approximate F F0 . E E = 2 3 4 3 2 2 3 F 3 2 π n E m = = . 3 28 3 26 2 3 4 3 34 2 28 3 2 3 19 F 31 851kg m But the electron concentration 1.31 10 electron m 6.49 10 kg 3 (1.054 10 J s) (1.31 10 /m ) 3.24 10 J 2.03 eV. 2(9.11 10 kg) ρ n n m π E = = = × × × × = = × = × 42.50. I DENTIFY : The only difference between the two isotopes is their mass, which will affect their reduced mass and hence their moment of inertia. S ET U P : The rotational energy states are given by 2 ( 1) 2 E l l I = + = and the reduced mass is given by m r = m 1 m 2 /( m 1 + m 2 ). E XECUTE : (a) If we call m the mass of the H-atom, the mass of the deuterium atom is 2 m and the reduced masses of the molecules are H 2 (hydrogen): m r (H) = mm/ ( m + m ) = m /2 D 2 (deuterium): m r (D) = (2 m )(2 m )/(2 m + 2 m ) = m Using I = m r r 0 2 , the moments of inertia are I H = mr 0 2 /2 and I D = mr 0 2 . The ratio of the rotational energies is then ( ) ( ) 2 2 H H D 0 2 2 D H D 0 ( 1) /2 2 ( 1) / 2 2 l l I E I mr m E I l l I r + = = = = + = = . (b) The ratio of the vibrational energies is r H r D r r 1 2 (H) (D) 2 (H) /2 1 2 (D) k n m E m m E m m k n m + = = = = + = = . E VALUATE : The electrical force is the same for both molecules since both H and D have the same charge, so it is reasonable that the force constant would be the same for both of them. 41.51. I DENTIFY and S ET U P : Use the description of the bcc lattice in Fig.42.12c in the textbook to calculate the number of atoms per unit cell and then the number of atoms per unit volume. E XECUTE : (a) Each unit cell has one atom at its center and 8 atoms at its corners that are each shared by 8 other unit cells. So there are 1 8/8 2 + = atoms per unit cell. 28 3 9 3 2 4.66 10 atoms/m (0.35 10 m) n V = = × × (b) 2 / 3 2 / 3 4 / 3 2 F0 3 2 N E m V π = = In this equation N / V is the number of free electrons per 3 m . But the problem says to assume one free electron per atom, so this is the same as n / V calculated in part (a). 31 9.109 10 kg m = × (the electron mass), so 19 F0 7.563 10 J 4.7 eV E = × = E VALUATE : Our result for metallic lithium is similar to that calculated for copper in Example 42.8.
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Molecules and Condensed Matter 42-13 42.52. (a) 2 tot 2 9 0 1 1 8 .
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