YF_ISM_38

YF_ISM_38 - PHOTONS ELECTRONS AND ATOMS 38 h f The e e 38.1...

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38-1 P HOTONS , E LECTRONS , AND A TOMS 38.1. IDENTIFY and SET UP: The stopping potential V 0 is related to the frequency of the light by 0 h Vf ee φ =− . The slope of V 0 versus f is h/e . The value f th of f when 0 0 V = is related to by th hf = . EXECUTE: (a) From the graph, 15 th 1.25 10 Hz f . Therefore, with the value of h from part (b), th 4.8 eV hf == . (b) From the graph, the slope is 15 3.8 10 V s ×⋅ . 16 15 34 ( )(slope) (1.60 10 C)(3.8 10 V s) 6.1 10 J s he −− × × = × (c) No photoelectrons are produced for th ff < . (d) For a different metal f th and are different. The slope is h/e so would be the same, but the graph would be shifted right or left so it has a different intercept with the horizontal axis. EVALUATE: As the frequency f of the light is increased above f th the energy of the photons in the light increases and more energetic photons are produced. The work function we calculated is similar to that for gold or nickel. 38.2. IDENTIFY and SET UP: cf λ = relates frequency and wavelength and Eh f = relates energy and frequency for a photon. 8 3.00 10 m/s c . 16 1 eV 1.60 10 J . EXECUTE: (a) 8 14 9 5.94 10 Hz 505 10 m c f × = × × (b) 34 14 19 (6.626 10 J s)(5.94 10 Hz) 3.94 10 J 2.46 eV f × × = × = (c) 2 1 2 Km v = so 19 15 22 ( 3 . 9 4 1 0 J ) 9.1 mm/s 9.5 10 kg K v m × = × 38.3. 8 14 7 3.00 10 m s 5.77 10 Hz λ 5.20 10 m c f × = × × 34 27 7 27 8 19 6.63 10 J s 1.28 10 kg m s λ 5.20 10 m (1.28 10 kg m s) (3.00 10 m s) 3.84 10 J 2.40 eV. h p Ep c = × × × × = × = 38.4. IDENTIFY and SET UP: av energy P t = . 19 1 eV 1.60 10 J . For a photon, hc f . 34 6.63 10 J s h . EXECUTE: (a) 32 1 6 av energy (0.600 W)(20.0 10 s) 1.20 10 J 7.5 10 eV Pt × = × = × (b) 34 8 19 9 (6.63 10 J s)(3.00 10 m/s) 3.05 10 J 1.91 eV 652 10 m hc E ×⋅ × = × = × (c) The number of photons is the total energy in a pulse divided by the energy of one photon: 2 16 19 1.20 10 J 3.93 10 photons 3.05 10 J/photon × × . EVALUATE: The number of photons in each pulse is very large. 38.5. IDENTIFY and SET UP: Eq.(38.2) relates the photon energy and wavelength. = relates speed, frequency and wavelength for an electromagnetic wave. EXECUTE: (a) f = so 61 9 20 34 (2.45 10 eV)(1.602 10 J/1 eV) 5.92 10 Hz 6.626 10 J s E f h ×× = × (b) = so 8 13 20 2.998 10 m/s 5.06 10 m c f × = × × (c) EVALUATE: is comparable to a nuclear radius. Note that in doing the calculation the energy in MeV was converted to the SI unit of Joules. 38
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38-2 Chapter 38 38.6. IDENTIFY and SET UP: th 272 nm λ = . cf = . 2 max 1 2 mv hf φ =− . At the threshold frequency, th f , max 0 v . 15 4.136 10 eV s h . EXECUTE: (a) 8 15 th 9 th 3.00 10 m/s 1.10 10 Hz 272 10 m c f × == = × × . (b) 15 15 th (4.136 10 eV s)(1.10 10 Hz) 4.55 eV hf × × = . (c) 21 5 1 5 max 1 (4.136 10 eV s)(1.45 10 Hz) 4.55 eV 6.00 eV 4.55 eV 1.45 eV 2 mv hf = = ×⋅ ×−=−= EVALUATE: The threshold wavelength depends on the work function for the surface. 38.7. IDENTIFY and SET UP: Eq.(38.3): 2 max 1 . 2 hc mv hf φφ =−= − Take the work function from Table 38.1. Solve for max . v Note that we wrote f as / . c EXECUTE: 34 8 2 19 max 9 1 (6.626 10 J s)(2.998 10 m/s) (5.1 eV)(1.602 10 J/1 eV) 2 235 10 m mv × × × 91 9 2 0 max 1 8.453 10 J 8.170 10 J 2.83 10 J 2 mv −− −× = × 20 5 max 31 2(2.83 10 J) 2.49 10 m/s 9.109 10 kg v × × × EVALUATE: The work function in eV was converted to joules for use in Eq.(38.3). A photon with
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YF_ISM_38 - PHOTONS ELECTRONS AND ATOMS 38 h f The e e 38.1...

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