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UVM PHYS 012 - Photoelectric and Compton Effects
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PHYSICS 012 1st Edition Lecture 32 Outline of Last Lecture I. Problem: A diverging lens with focal length of -20 cm is placed 50 cm to the left of a converging lens with focal length of 15 cm. An object is placed 10 cm to the left of the diverging lens. How far is the final image from the second lens?a. 1/di1 = 1/f1 – 1/do1 = 1/-20 – 1/10 = -3/20b. di1 = -20/3 cmc. do2 = di1 + 50 cmd. 1/di2 = 1/f2 – 1/do2 = 1/15 – 1/(50 + di1)III. Problem 47 from chapter 26 in text booka. nred = 1.662b. nairsinθ1 = nredsinθ2c. θ2 = sin-1(sin60/nred) = 31.4 degreesd. θ3 = 90 – (180-60-58.6) = 90 – 61.4 = 28.6 degreese. nredsinθ3 = nairsinθ4f. θ4 = sin-1(nredsin(28.6)) = 52.7 degreesIV. Problem 37 from chapter 27 in text bookThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.a. λ = 498 nm, L = 1.6 kmb. θmin = 1.22λ/Dc. sinθ ≈ h/L ≈ θd. D = 1.22λ/θmin = 1.22λL/h = 0.0097 mOutline of Current Lecture V. Photoelectric Effecta. Light ejects electrons from the back metal plate to create a current. Increasing voltage increases current.b. Observed Effectsi. Maximum kinetic energy of electrons is independent of light intensity.ii. Electrons are emitted instantaneously, even at lower intensity.iii. No electrons are emitted below “cutoff frequency,” regardless of intensity.1. fco = Wo/h2. Wo = work function of metal (like binding energy, varies based on material)iv. Maximum possible kinetic energy increases with increasing frequency.1. E = hf = KEmax + Wo2. hc = 1240 eV*nmVI. Compton Effecta. When a photon hits a stationary electron, both the electron and the photon are scattered. The photon has a new frequency.i. hf = hf` + KEelectronii. λ + λ` = (h/mec) (1-cosθ)iii. 0° < θ < 180°iv. 0 nm < (λ – λ`) < 2h/mec


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UVM PHYS 012 - Photoelectric and Compton Effects

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