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Chapter 39

# Chapter 39 - 39.1 Solve A steady photoelectric current of...

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39.1. Solve: A steady photoelectric current of 10 μ A is indicated in the graph. The number of electrons per second is 5 13 19 C C 1 electron 10 A 10 1.0 10 6.25 10 electrons/s s s 1.6 10 C μ μ = = × × = × ×

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39.2. Model: Light of frequency f consists of discrete quanta, each of energy E = hf . Solve: (a) The energy of the light quantum is ( )( ) 34 8 9 19 6.63 10 J s 3.0 10 m/s 1 eV 3.11 eV 400 10 m 1.6 10 J c E hf h λ × × = = = × = × × From Table 39.1, the work functions for sodium and potassium are smaller than 3.11 eV. That is, light of wavelength 400 nm has enough energy to eject photoelectrons from sodium and potassium. (b) The energy of the light quantum is ( )( ) 34 8 9 19 6.63 10 J s 3.0 10 m/s 1 eV 4.97 eV 250 10 m 1.6 10 J c E hf h λ × × = = = × = × × Light of wavelength 250 nm has enough energy to eject photoelectrons from all of the metals on the table except gold.
39.3. Model: Light of frequency f consists of discrete quanta, each of energy E = hf . Solve: The lowest photon energy that creates photoelectrons from the metal is ( )( ) 34 8 9 19 6.63 10 J s 3.0 10 m/s 1 eV 3.20 eV 388 10 m 1.6 10 J hc E λ × × = = × = × × The work function of the metal is E 0 = 3.20 eV.

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39.4. Solve: From Equation 39.7, the maximum kinetic energy is ( )( ) max 0 0 34 8 19 0 max 6.63 10 J s 3.0 10 m/s 1 eV 209 nm 4.65 eV 1.30 eV 1.6 10 J c K hf E h E hc E K λ λ = = × × = = × = + + × Assess: λ = 209 nm is the wavelength of light in the ultraviolet region of the spectrum.
39.5. Model: The threshold frequency for the ejection of photoelectrons is f 0 = E 0 / h where E 0 is the work function. Solve: The visible region of light extends from 400 nm to 700 nm. For λ 0 = 400 nm, the work function is ( )( ) 34 8 0 0 9 19 0 6.63 10 J s 3.0 10 m/s 1 eV 3.11 eV 400 10 m 1.6 10 J hc E f h λ × × = = = × = × × For λ 0 = 700 nm, ( )( ) 34 8 0 9 19 6.63 10 J s 3.0 10 m/s 1 eV 1.78 eV 700 10 m 1.6 10 J E × × = × = × × The cathode that will work in the entire visible range must have a work function of 1.78 eV or less.

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39.6. Visualize: Please refer to Figure 39.10. Solve: (a) The threshold frequency is seen to be f 0 = 4.39 × 10 14 Hz. According to Einstein’s theory of the photoelectric effect, the work function is E 0 = hf 0 . Using the modern value of h , the work function of cesium is ( )( ) 15 14 0 0 4.14 10 eV s 4.39 10 Hz 1.82 eV E hf = = × × = (b) According to Einstein’s theory, a graph of V stop versus f should be linear with a slope of h / e . Millikan’s data is seen to be linear with an experimental slope of 4.124 × 10 15 V/Hz. So, an experimental value of h is h = e × slope = (1.60 × 10 19 C)(4.124 × 10 15 V/Hz) = 6.60 × 10 34 J s
39.7. Solve: (a) A metal can be identified by its work function. From Equation 39.8, the stopping potential is 0 stop hf E V e = E 0 = hf – eV stop The frequency and energy of the photons are 8 15 9 3.00 10 m/s 1.500 10 Hz 200 10 m c f λ × = = = × × ( )( ) 15 15 4.14 10 eV s 1.500 10 Hz 6.21 eV hf = × × = If the stopping potential is V stop = 1.93 V, then eV stop = 1.93 eV. Thus, 0 stop 6.21 eV 1.93 eV 4.28 eV E hf eV = = = Using Table 39.1, we can identify the metal as aluminum .

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