39 1 Solve A steady photoelectric current of 10 A is indicated in the graph The number of electrons per second is 10 A 10 1 0 10 5 6 25 10 electrons s 13 C s C s 1 electron 1 6 10 19 C 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 E hf h c 6 63 10 J s 3 0 10 m s 8 34 400 10 m 9 1 eV 1 6 10 19 J 3 11 eV 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 J s 3 0 10 m s 6 63 10 8 E hf h c 34 250 10 m 9 1 eV 1 6 10 19 J 4 97 eV 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 E hc 6 63 10 J s 3 0 10 m s 8 34 388 10 m 9 1 eV 1 6 10 19 J 3 20 eV The work function of the metal is E0 3 20 eV 39 4 Solve From Equation 39 7 the maximum kinetic energy is K max hf E 0 h E 0 c 8 hc E K 0 max 6 63 10 34 J s 3 0 10 m s 4 65 eV 1 30 eV 1 eV 1 6 10 19 J 209 nm Assess 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 f0 E0 h where E0 is the work function Solve The visible region of light extends from 400 nm to 700 nm For 0 400 nm the work function is 400 10 m 1 eV 1 6 10 J s 3 0 10 m s 6 63 10 3 11 eV f h 0 E 0 J 9 34 19 8 hc 0 For 0 700 nm E 0 6 63 10 J s 3 0 10 m s 8 34 700 10 m 9 1 eV 1 6 10 19 J 1 78 eV The cathode that will work in the entire visible range must have a work function of 1 78 eV or less 39 6 Visualize Please refer to Figure 39 10 Solve photoelectric effect the work function is E0 hf0 Using the modern value of h the work function of cesium is a The threshold frequency is seen to be f0 4 39 1014 Hz According to Einstein s theory of the b According to Einstein s theory a graph of Vstop 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 eV s 4 39 10 Hz 4 14 10 1 82 eV hf E 15 14 0 0 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 V stop hf E 0 e E0 hf eVstop The frequency and energy of the photons are f c 8 3 00 10 m s 200 10 m 9 1 500 10 Hz hf 4 14 10 eV s 1 500 10 Hz 6 21 eV 15 15 15 If the stopping potential is Vstop 1 93 V then eVstop 1 93 eV Thus E 0 hf eV stop 6 21 eV 1 93 eV 4 28 eV Using Table 39 1 we can identify the metal as aluminum b The kinetic energy of the electrons and thus the stopping potential are independent of the light intensity A more intense light generates more electrons but the electrons still have the same kinetic energy The stopping potential is Vstop 1 93 V after the intensity is doubled 39 8 Solve a The frequency of the photon is f c 8 3 00 10 m s 700 10 m 9 4 29 10 Hz 14 From Equation 39 4 the energy is b The frequency of the photon is Thus the wavelength is E hf 4 14 10 eV s 4 29 10 Hz 1 77 eV 15 14 f E h 5000 eV 4 14 10 15 eV s 1 208 10 Hz 18 c f 8 3 00 10 m s 1 208 10 Hz 18 2 5 10 10 m 0 25 nm Assess Because x ray photons are very energetic their wavelength is small 39 9 Solve a Using Equation 39 4 the wavelength of the photon is c f c E h hc E 4 14 10 15 eV s 3 10 m s 8 0 30 eV 4 14 10 m 4140 nm 6 This is infrared light b Likewise for an energy of 3 0 eV the wavelength is 414 nm and is in the visible region c For an energy of 30 0 eV the wavelength is 41 4 nm and is in the ultraviolet region Assess Since E 1 the higher the energy of the photon the smaller its wavelength 39 10 Solve a From Equation 39 4 the energy of the radio frequency photon is E hf 4 14 10 15 eV s 100 10 Hz 6 4 14 10 eV 7 b The energy of the visible light photon hc c The energy of the x ray photon is E 4 14 10 eV s 3 0 10 m s 8 15 500 10 m 9 2 49 eV E hc 4 14 10 eV s 3 0 10 m s 8 15 0 10 10 m 9 12 4 keV 39 11 Solve a From Equation 39 4 the energy of each photon is Ephoton hf 6 63 10 34 J s 101 106 Hz 6 696 10 28 J The number of photons in 104 J is N total E E photon 4 10 J 6 696 10 28 J 1 5 10 29 The antenna emits 1 5 1029 photons per second b The number of photons emitted per second is so enormous that we couldn t possibly recognize the effects of single photons It s safe to treat the broadcast as an electromagnetic wave 39 12 Solve The rate of photon emission by a laser is the number of photons emitted per second If P is the power of the beam the rate is R photon P hf P hc P hc R R red blue P red hc hc P blue red blue …
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