1116_PartUniversity Physics Solution

1116_PartUniversity Physics Solution - 41-10 41.51. Chapter...

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41-10 Chapter 41 45 5 and 1.62 10 eV 8.11 10 eV 8.09 10 eV from part (a). E −− − Δ= × × = × Then, 11 | | 2.81 10 m 0.0281 nm λ × = . The wavelength corresponds to a larger energy change, and so the wavelength is smaller. 41.51. IDENTIFY: The ratio according to the Boltzmann distribution is given by Eq.(38.21): 10 () / 1 0 EE k T n e n −− = , where 1 is the higher energy state and 0 is the lower energy state. SET UP: The interaction energy with the magnetic field is 2.00232 2 zs e UB m B m μ ⎛⎞ =− = ⎜⎟ ⎝⎠ = (Example 41.5.). The energy of the 1 2 s m =+ level is increased and the energy of the 1 2 s m level is decreased. 1/2 / UU k T n e n = EXECUTE: B 11 2.00232 2.00232 2.00232 22 2 2 ee B B B mm −= = = == B (2.00232) / Bk T n e n = (a) 5 5.00 10 T B 24 2 5 23 2.00232(9.274 10 A/m )(5.00 10 T)/([1.381 10 J/K][300 K]) n e n −× × × = 7 2.24 10 7 0.99999978 1 2.2 10 n e n = × (b) 3 52 . 2 4 1 0 5.00 10 T, 0.9978 n Be n × = = (c) 2 . 2 4 1 0 0.978 n n × = = EVALUATE: For small fields the energy separation between the two spin states is much less than kT for 300 K T = and the states are equally populated. For 5.00 T B = the energy spacing is large enough for there to be a small excess of atoms in the lower state. 41.52. Using Eq.(41.4), (1 ) , Lm v r l l ==+ = and the Bohr radius from Eq.(38.15), we obtain the following value for v : 34 5 23 1 1 1 0 ) 2(6.63 10 J s) 7.74 10 m s. ( ) 2 (9.11 10 kg) (4) (5.29 10 m) ll v mna π + ×⋅ = × ×× = The magnetic field generated by the “moving” proton at the electrons position can be calculated from Eq.(28.1): 19 5 7 0 1 1 2 | | sin (1.60 10 C) (7.74 10 m s) sin(90 ) (10 T m A) 0.277 T. 4( 4 ) ( 5 . 2 9 1 0 m ) μ qv B π r φ ° = × 41.53. IDENTIFY and SET UP: s m can take on 4 different values: 3113 , . 2222 s m + + Each l nlm state can have 4 electrons, each with one of the four different s m values. Apply the exclusion principle to determine the electron configurations. EXECUTE: (a) For a filled 1 n = shell, the electron configuration would be 4 1; s four electrons and 4. Z = For a filled 2 n = shell, the electron configuration would be 44 1 2 122 ; ss p twenty electrons and 20. Z = (b) Sodium has 11; Z = 11 electrons. The ground-state electron configuration would be 44 3 122 . EVALUATE: The chemical properties of each element would be very different. 41.54. (a) ( 13.6 eV) (7) ( 13.6 eV) 666 eV. Z (b) The negative of the result of part (a), 666 eV. (c) The radius of the ground state orbit is inversely proportional to the nuclear charge, and 10 12 (0.529 10 m) 7 7.56 10 m. a Z = × (d) 0 12 hc hc E E Δ , where 0 E is the energy found in part (b), and 2.49 nm. =
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Atomic Structure 41-11 41.55. (a) IDENTIFY and SET UP: The energy of the photon equals the transition energy of the atom: / . Eh c λ Δ= The energies of the states are given by Eq.(41.3). EXECUTE: 2 13.60 eV n E n =− so 2 13.60 eV 4 E and 1 13.60 eV 1 E 19 18 21 13 13.60 eV 1 (13.60 eV) 10.20 eV (10.20 eV)(1.602 10 J/eV) 1.634 10 J 44 EE E −− ⎛⎞ Δ= − = −+ = = = × = × ⎜⎟ ⎝⎠ 34 8 7 18 (6.626 10 J s)(2.998 10 m/s) 1.22 10 m 122 nm 1.634 10 J hc E ×⋅ × == = Δ× (b) IDENTIFY and SET UP: Calculate the change in E Δ due to the orbital magnetic interaction energy, Eq.(41.17), and relate this to the shift Δ in the photon wavelength.
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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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1116_PartUniversity Physics Solution - 41-10 41.51. Chapter...

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