Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

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Unformatted text preview: that the photon lies in the infrared region of the electromagnetic spectrum. ______________________________________________________________________________ 49. REASONING The number of photons is the energy delivered to the iris divided by the energy per photon. The energy E of a photon is given by Equation 29.2 as E = hf, where h is Planck’s constant and f is the photon frequency. The frequency is related to the wavelength λ by f = c/λ, according to Equation 16.1, where c is the speed of light in a vacuum. Substituting this expression for f into the expression for E gives E = hf = hc (1) λ SOLUTION Using Equation (1), we find that the number of photons is ( ) 4.1×10−3 J λ 4.1×10−3 J 4.1×10−3 J Number of photons = = = E hc / λ hc ( 4.1×10−3 J )(1064 ×10−9 m ) = 2.2 ×1016 = ( 6.63 ×10−34 J ⋅ s )(3.00 ×108 m/s ) 50. REASONING AND SOLUTION For either laser, the number of photons is given by the energy E produced divided by the energy per photon. Since power P is energy per unit time, the energy produced in a time t is E = Pt. For a photon frequency f the energy per photon is hf. But the photon frequency f and wavelength λ are related to the speed of light c by f = c/λ. Therefore, the energy per photon is hc/λ. The number of photons is, then, Number of photons = Energy produced Pt Pt λ = = Energy per photon hc / λ hc Since each laser produces the same number of photons, it follows that Pt λ Pt λ = hc He/Ne hc SS or tHe/Ne = ( Pt λ )SS ( Pλ )He/Ne 1576 THE NATURE OF THE ATOM where the "He/Ne" and "SS" refer to the helium/neon and solid state lasers, respectively. The time for the helium/neon laser is, then, ( Pt λ )SS tHe/Ne = ( Pλ )He/Ne ( (1.0 × 1014 W ) (1.1 × 10–11 s ) (1060 nm ) = 1.8 × 106 s = (1.0 × 10–3 W ) ( 633 nm ) ) ( ) 6 4 This is a time of 1.8 × 10 s (1 day ) / 8.64 × 10 s = 21 days . ______________________________________________________________________________ 51. REASONING AND SOLUTION According to Figure 30.15, the energy sublevel with n = 4, l = 0 (which corresponds to the notation 4s ) is lower in energy than the n = 3, l = 2 (which corresponds to the notation 3d ) energy sublevel. Thus, the 4s energy sublevel will be filled before the 3d energy sublevel. Therefore, using Figure 30.15 as a guide, we find that the ground state electronic configuration of arsenic (Z = 33) is 1s2 2s2 2p 6 3s2 3p6 4s2 3d10 4p3 ______________________________________________________________________________ 52. REASONING AND SOLUTION The first excited state occurs for n = 2 and has an energy given by Equation 30.13: E2 = – (13.6 eV ) Z2 n2 = – (13.6 eV ) 12 22 = –3.40 eV After absorbing 2.86 eV of energy, the electron has an energy En = –3.40 eV + 2.86 eV = –0.54 eV Again using Equation 30.13, we find En = – 0.54 eV = – (13.6 eV)(12/n2). Solving for n, we obtain 13.6 eV n= =5 0.54 eV ______________________________________________________________________________ 53. SSM REASONING In the theory of quantum mechanics, there is a selection rule that restricts the initial and final values of the orbital quantum number l . The selection rule states that when an electron makes a transition between energy levels, the value of l may not remain the same or increase or decrease by more than one. In other words, the rule requires that ∆l = ±1 . SOLUTION Chapter 30 Problems 1577 a. For the transition 2s → 1s , the electron makes a transition from the 2s state ( n = 2, l = 0 ) to the 1s state ( n = 1, l = 0 ) . Since the value of l is the same in both states, ∆l = 0 , and we can conclude that this energy level transition is not allowed . b. For the transition 2p → 1s , the electron makes a transition from the 2p state ( n = 2, l = 1) to the 1s state ( n = 1, l = 0 ) . The value of l changes so that ∆l = 0 − 1 = −1 , and we can conclude that this energy level transition is allowed . c. For the transition 4p → 2p , the electron makes a transition from the 4p state ( n = 4, l = 1) to the 2p state ( n = 2, l = 1) . Since the value of l is the same in both states, ∆l = 0 , and we can conclude that this energy level transition is not allowed . d. For the transition 4s → 2p , the electron makes a transition from the 4s state ( n = 4, l = 0 ) to the 2p state ( n = 2, l = 1) . The value of l changes so that ∆l = 1 − 0 = +1 , and we can conclude that this energy level transition is allowed . e. For the transition 3d → 3s , the electron makes a transition from the 3d state ( n = 3, l = 2) to the 3s state ( n = 3, l = 0 ) . The value of l changes so that ∆l = 0 − 2 = −2 , and we can conclude that this energy level transition is not allowed . ______________________________________________________________________________ 54. REASONING AND SOLUTION From Equation 30.14 we have 1 2 1 = R ( Z –1) 2 – 2 n λ f ni 1 Solving for Z we obtain Z= 1/ ( λ R ) (1/ nf2 ) – (1/ ni2 ) +1= ( )( )( ) 1/ 4.5 × 10 –9 m 1.097 × 107 m –1 + 1 = 6.2 2 2 1/1 – 1/ 2 ( ) Therefore, the element is likely to be...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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