1131_PartUniversity Physics Solution

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

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42-12 Chapter 42 (ii) transition 10 , 2 1 nn l l =→ = = →= 22 2 (6 2) 2 Eh f h f II ⎛⎞ Δ= − + = + ⎜⎟ ⎝⎠ == 8 11 13 2.998 10 m/s 4.28 m 2( /2 ) 2(3.960 10 Hz) 6.93 10 Hz c If λ μ π × = + × = (iii) transition 21 , 23 ll =→= 3 (6 12) 2 f h f + 8 11 13 4.40 m 3( /2 ) 3(3.960 10 Hz) c × = −+ × + × = EVALUATE: The vibrational energy change for the 1 0 transition is the same as for the 2 1 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 / FF 111 () 1 . 11 1 1 Ek T E kT E kT E kT E kT e fE E E ee e e −Δ Δ + = + = + = ++ + + 42.49. Since potassium is a metal we approximate 0 . EE = 23 43 2 23 F 3 2 π n E m = = . 3 28 3 26 23 43 34 2 28 3 23 19 F 31 851kg m But the electron concentration 1.31 10 electron m 6.49 10 kg 3( 1 . 0 5 4 1 0 J s ) ( 1 . 3 1 1 0 / m ) 3.24 10 J 2.03 eV. 2(9.11 10 kg) ρ m π E = × × ×⋅ × × = × 42.50. IDENTIFY: The only difference between the two isotopes is their mass, which will affect their reduced mass and hence their moment of inertia. SET UP: The rotational energy states are given by 2 (1 ) 2 El l I =+ = and the reduced mass is given by m r = m 1 m 2 /( m 1 + m 2 ). EXECUTE: (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 HD 0 2 2 DH D 0 ) / 2 2 2 2 I EI m r m I r + = = + = = . (b) The ratio of the vibrational energies is r Hr Dr r 1 2( H ) (D) 2 (H) /2 1 D ) k n m Em m m k n m + = = + = = . EVALUATE: 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. IDENTIFY and SET UP: 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. EXECUTE: (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 93 2 4.66 10 atoms/m (0.35 10 m) n V × × (b) 2/3 4/3 2 F0 3 2 N E mV = = 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 = EVALUATE: 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 29 0 11 8. 4 d α e UA dr π rr =− P Setting this equal to zero when 0 = gives 7 0 0 2 84 A r α e π = P and so 27 0 tot 8 0 1 . 48 α er U π ⎛⎞ + ⎜⎟ ⎝⎠ P At 0 , = 2 18 tot 00 7 1.26 10 J 7.85 eV. 32 α e U π r × P (b) To remove a Na Cl +− ion pair from the crystal requires 7.85 eV. When neutral Na and Cl atoms are formed from the Na + and Cl atoms there is a net release of energy 5.14 eV 3.61eV 1.53 eV, −+ = so the net energy required to remove a neutral Na, Cl pair from the crystal is 7.85 eV 1.53 eV 6.32 eV.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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

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