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# f02x3aan - 8 m a a = 6.9 ×10-7 m = 0.69 m c 2 8 31 2 34 2...

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NAME PHYS-1200 PHYSICS II QUIZ 3 DECEMBER 4, 2002 SOLUTIONS PART A. 1. E 6. C 2. A 7. D 3. D 8. A 4. D 5. B PART B . 1. a) The answer can be read off the graph, where the stopping potential equals zero. f c = 3.0 ×10 14 Hz b) When the stopping potential equals zero, 0 = hf c Φ , so Φ = hf c . Φ = (6.63 ×10 -34 J·s)(3.0 ×10 14 Hz) Φ = 2.0 ×10 -19 J = 1.2 eV 2. a) E g = 20 eV – 13 eV E g = 7 eV b) The Fermi energy is the energy of the highest filled state, measured from the bottom of the conduction band . In this case, that is 25 eV – 20 eV. E F = 5 eV c) CONDUCTOR . The highest occupied band is only partially filled.

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NAME PART C . 1. a) d sin θ = m λ , so ° × = = - sin1.0 m) 10 0 . 6 ( 1 sin 8 m d d = 3.4 ×10 -6 m = 3.4 μ m b) a sin = m , so ° × = = - sin5.0 m) 10 0 . 6 ( 1 sin
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Unformatted text preview: 8 m a a = 6.9 ×10-7 m = 0.69 m c) 2 8 31 2 34 2 2 2 m) 10 kg)(6.0 10 2(9.11 s) J 10 63 . 6 ( 2 2---× × ⋅ × = = = m h m p E el E el = 6.7 ×10-23 J = 4.2 ×10-4 eV = 0.42 meV e) ∆ y = D tan = (0.50 m) tan 1.0° ∆ y = 8.7 ×10-3 m = 8.7 mm 2. a) b) Greater than 2.0 nm For an infinitly deep well, the wavelength would be double the width of the well. For a finite well, the wavelength must be greater. c) m 10 620 m/s) 10 . 3 s)( J 10 63 . 6 ( 9 8 34--× × ⋅ × = = = hc hf E ph E ph = 3.2 ×10-19 J = 2.0 eV d) E 3 = 2.27 eV E 1 + E ph = 0.27 eV + 2.0 eV = 2.27 eV = E 3...
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f02x3aan - 8 m a a = 6.9 ×10-7 m = 0.69 m c 2 8 31 2 34 2...

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