Physics Solution Manual for 1100 and 2101

3 the binding energy is the energy that must be

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Unformatted text preview: )( Z − 1) 2 (1) 1572 THE NATURE OF THE ATOM We can use Equation 30.14 to estimate the Kα wavelength λ by recognizing that an electron in the L shell (ni = 2) falls into the K shell (nf = 1) when the Kα photon is emitted. In addition, we use Z − 1 instead of Z to account for shielding (see Example 11). Thus, we have 1 1 2 1 2 1 = R ( Z − 1) 2 − 2 = R ( Z − 1) 2 − 2 n λ 1 2 f ni 1 (2) According to Equation (2), the wavelength λ is inversely proportional to (Z − 1)2. Therefore, a larger value of the wavelength implies a smaller value of Z − 1. If Z − 1 is smaller, then Z must also be smaller. Thus, since λA is greater than λB, it follows that ZA is less than ZB. According to Equation (1), Emin is directly proportional to (Z − 1)2. This means that when Z is larger, Emin is also larger. However, we have just seen that ZA is less than ZB, so we conclude that Emin, A is less than Emin, B. SOLUTION Applying Equation (1) to both metals A and B, we have Emin, A Emin, B (13.6 eV ) ( Z A − 1) ( Z A − 1) = = 2 2 (13.6 eV ) ( Z B − 1) ( Z B − 1) 2 2 (3) According to Equation (2), the wavelength λα and (Z − 1)2 are inversely proportional. Thus, Equation (3) becomes Emin, A Emin, B = 1/ λA 1/λB = λB λA (4) Since λA = 2λB , we have Emin, A Emin, B = λB λB = = 0.50 λA 2λB As expected Emin, A is less than Emin, B. ______________________________________________________________________________ 45. SSM WWW REASONING The number of photons emitted by the laser will be equal to the total energy carried in the beam divided by the energy per photon. SOLUTION The total energy carried in the beam is, from the definition of power, E total = Pt = (1.5 W)(0.050 s) = 0.075 J Chapter 30 Problems 1573 The energy of a single photon is given by Equations 29.2 and 16.1 as Ephoton = hf = hc λ ( 6.63 ×10−34 J ⋅ s ) (3.00 ×108 m/s ) = 3.87 ×10−19 J = 514 × 10−9 m where we have used the fact that 514 nm = 514 × 10–9 m . Therefore, the number of photons emitted by the laser is Etotal Ephoton = 0.075 J = 1.9 × 1017 photons −19 3.87 × 10 J/photon ______________________________________________________________________________ 46. REASONING The total energy Etot of a single pulse is equal to the number N of photons in the pulse multiplied by the energy E of each photon: Etot = NE (1) The energy E of each photon is given by E = hf (Equation 29.2), where h = 6.63×10−34 J·s c is Planck’s constant and f is the frequency of the photon. We will use f = λ (Equation 16.1) to determine the frequency f of the photons from their wavelength λ and the speed c of light in a vacuum. To find the total energy Etot of each pulse, we will make use of the fact that the average power Pavg of the laser is equal to the total energy of a single pulse divided by the duration ∆t of the pulse: Pavg = Etot ∆t (6.10b) SOLUTION Solving Equation (1) for N, we obtain N= Etot (2) E Solving Equation (6.10b) for Etot yields Etot = Pavg ∆t . Substituting this result and E = hf (Equation 29.2) into Equation (2), we find that N= Substituting f = each pulse is: c λ Etot E = Pavg ∆t hf (3) (Equation 16.1) into Equation (3), we find that the number of photons in 1574 THE NATURE OF THE ATOM N= Pavg ∆t hf = Pavg ∆t c h λ = Pavg ∆tλ hc = ( 5.00 ×10−3 W ) ( 25.0 ×10−3 s ) ( 633 ×10−9 m ) ( 6.63 ×10−34 J ⋅ s )( 3.00 ×108 m/s ) = 3.98 ×1014 47. REASONING AND SOLUTION According to Equations 29.2 and 16.1, E = hc / λ , the energy of a single photon of the dye laser is E= hc λ = (6.63 ×10 –34 J ⋅ s)(3.00 ×108 m/s) = 3.40 ×10 –19 J –9 585 ×10 m while the energy of a single photon from the CO 2 laser is E= hc λ = (6.63 ×10 –34 J ⋅ s)(3.00 ×108 m/s) = 1.88 ×10 –20 J –5 1.06 ×10 m The ratio of these two photon energies is 3.40 × 10–19 J 1.88 × 10 –20 J = 18.1 We can only have a whole number of photons, therefore, the minimum number of photons that the carbon dioxide laser must produce to deliver at least as much energy to a target as does a single photon from the dye laser is 19 . ______________________________________________________________________________ 48. REASONING The external source of energy must “pump” the electrons from the ground state E0 to the metastable state E2. The population inversion occurs between the metastable state E2 and the one below it, E1. The lasing action occurs between the two states that have the population inversion, the E2 and E1 states. SOLUTION a. From the drawing, we see that the energy required to raise an electron from the E0 state to the E2 state is 0.289 eV . b. The lasing action occurs between the E2 and E1 states, and so the energy E of the emitted photon is the difference between them; E = E2 – E1. According to Equations 29.2 and 16.1, the wavelength λ of the photon is related to its energy via λ = hc/E, so that Chapter 30 Problems hc λ= = E2 − E1 ( 6.63 × 10−34 J ⋅ s ) ( 3.00 × 108 m/s ) 1.60 × 10 ( 0.289 eV − 0.165 eV ) −19 J 1 eV 1575 = 1.00 × 10−5 m c. An examination of Figure 24.9 shows...
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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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