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Recitation 5
Chapter 28 Problem 28.4. Calculate the energy, in electron volts, of a photon whose frequency is (a) 620 THz, (b) 3.10 GHz, and (c) 46.0 MHz. (d) Determine the corresponding wavelengths for these photons and state the classification of each on the electromagnetic spectrum. (a) 2.56 eV 484 nm Visible (blue) (b) 12.8 µeV 9.67 cm Microwave (c) 190 neV 6.51 m Radio

E = hf λ = c/fClass

See Figure 24.12 (page 823) in the text for a classification spectrum. Problem 28.6. The average threshold of dark-adapted (scotopic) vision is 4.00 · 10−11 W/m2 at a central wavelength of 500 nm. If light having this intensity and wavelength enters the eye and the pupil is open to its maximum diameter of 8.50 mm, how many photons per second enter the eye? The total power into the eye is P =IA = πr2 I = and the energy per photon is E = hf = so the number of photons entering per second is Φp = P = 5710 . E (3) hc = 2.48 eV = 3.97 · 10−19 J , λ (2) πd2 I = 2.27 fW , 4 (1)

Problem 28.9. Molybdenum has a work function of 4.20 eV. (a) Find the cutoff wavelength and cutoff frequency for the photoelectic effect. (b) What is the stopping potential if the incident light has a wavelength of 180nm? (a) The cutoff wavelength is the wavelength where the incoming light has barely enough energy to free an electron, i.e. all of the photon’s energy goes into overcoming the work function barrier. hf = φ φ 4.20 eV · 1.60 · 10 J/eV = = 1.01 PHz h 6.63 · 10−34 J·s λf = c c λ = = 296 nm f f=
−19

(4) (5) (6) (7)

(b) The photon brings in hf , but much of that energy goes to overcoming the workfunction barrier. The left over energy hf − φ becomes the electron’s kinetic energy. The stopping potential is the voltage change which matches that kinetic energy. Kmax = hf − φ = h c − φ = 4.30 · 10−19 J = 2.69 eV λ ∆VS = Kmax /e = 2.69 V (8) (9)

Problem 28.10. Electrons are ejected from a metallic surface with speeds ranging up to 4.60 · 105 m/s when light with a wavelength of 625 nm isused. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface? (a) Conserving energy hf = hc 1 2 = φ + Kmax = φ + me vmax (10) λ 2 hc 1 6.63 · 10−34 J·s · 3.00 · 108 m/s 1 2 φ= − me vmax = − · 9.11 · 10−31 kg · (4.60 · 105 m/s)2 = 2.21 · 10−19 J = 1.38 eV λ 2 625 · 10−9 m 2 (11)

(b) Light at the cutoff frequency only barely supplies enough energy to overcomethe work function. hfcut = φ φ fcut = = 334 THz h (12) (13) (14)

Problem 28.13. An isolated copper sphere of radius 5.00 cm, initially uncharged, is illuminated by ultraviolet light of wavelength 200 nm. What charge will the photoelectric effect induce on the sphere? The work function for copper is 4.70 eV. As light lands on the sphere, electrons are blasted off into oblivion. As they leave, thesphere accumulates positive charge, so it takes a bit more energy to pull the next electrons away. Eventually the system reaches equilibrium when the initial kinetic energy of electron blasting off is not quite enough for it to escape the electro-static attraction to the positive sphere. In math Kmax = hf − φ = = ke Q= Qq R hc − φ = 2.40 · 10−19 J = 1.50 eV λ (15) (16) (17)

Kmax R 2.40 · 10−19 J= 8.34 pC = 52.1 · 106 electrons = 9 N·m2 /C2 · 1.60 · 10−19 C ke q 8.99 · 10

Problem 28.56. Figure P28.56 shows the stopping potential versus the incident photon frequency for the photoelectric effect for sodupm. Use the graph to find (a) the work function, (b) the ratio h/e, and (c) the cutoff wavelength. The data are taken from R.A. Millikan, Physical Review 7:362 (1916).

4 ∆Vs (V) 3 2 1 0400 800 f (THz) 1200

(a) The fit line passes nearby the points (400 THz, 0 V) and (1.20 PHz, 3.3 V). In point-slope form, the fit line is then ∆Vs − 0 V = (3.3 − 0) V (f − 400 THz) (1200 − 400) THz ∆Vs = 4.125mV/THz · (f − 400 THz) (18) (19)

The work function is the inverse y-intercept, so φ = −∆Vs (f = 0) = −4.125mV/THz · (−400 THz) = 1.65 eV (b) The theoretical form for the fit line is e∆Vs...
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