# Lect09 - Le cture9 I ntroduction to QM Re w and Exam s vie ple S1 S2 Lecture 9 p 1 Photoe ctric Effe le ct KEmax = e Vstop = hf The work function q

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Lecture 9, p 1 Lecture 9: Introduction to QM: Review and Examples S 1 S 2

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Lecture 9, p 2 The work function: Φ is the minimum energy needed to strip an electron from the metal. Φ is defined as positive . Not all electrons will leave with the maximum kinetic energy (due to losses). Photoelectric Effect max stop KE e V hf = = - Φ f (x10 14 Hz) V stop (v) 0 0.5 1 1.5 2 2.5 3 3.5 0 5 10 15 Binding energy Φ Conclusions: Light arrives in “packets” of energy ( photons ). E photon = hf Increasing the intensity increases # photons, not the photon energy. Each photon ejects (at most) one electron from the metal. Recall: For EM waves, frequency and wavelength are related by f = c/ λ . Therefore: E photon = hc/ λ = 1240 eV-nm/ λ
Lecture 9, p 3 Act 1 1. If Planck’s constant were somewhat larger, but Φ remained the same, how would the graph change? a. Increased slope b. Increased f o c. Both a and b 2. If Φ increased, but h remained the same, how would the graph change? a. Increased slope b. Increased f o c. Both a and b V stop (v) f (x10 14 Hz) 0 1 2 3 0 5 10 15

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Lecture 9, p 4
Lecture 9, p 5 Summary: Light (v = c) E = pc, so E = hc/ λ Slow Matter (v << c) KE = p 2 /2m, so KE = h 2 /2m λ 2 For electrons: 1240 eV nm photon E λ = 2 2 1.505 eV nm KE = Everything E = hf p = h/ λ

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Act 2: Counting photons Which emits more photons , a 1-mW cell phone (f = 830 MHz λ = 0.36 m ) or a 1-mW laser ( λ = 635 nm )?
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## This note was uploaded on 02/21/2011 for the course PHYS 214 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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Lect09 - Le cture9 I ntroduction to QM Re w and Exam s vie ple S1 S2 Lecture 9 p 1 Photoe ctric Effe le ct KEmax = e Vstop = hf The work function q

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