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Unformatted text preview: Lecture 9, p 1 Lecture 9, p 2 Lecture 9: Introduction to QM: Review and Examples S 1 S 2 Lecture 9, p 3 The work function: c is the minimum energy needed to strip an electron from the metal. c is defined as positive . c Not all electrons will leave with the maximum kinetic energy (due to losses). Photoelectric Effect max stop KE e V hf = = f (x10 14 Hz) V stop (v) 0.5 1 1.5 2 2.5 3 3.5 5 10 15 f Binding energy 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/ . Therefore: E photon = hc/ = 1240 eVnm/ Lecture 9, p 4 Act 1 1. If Plancks constant were somewhat larger, but remained the same, how would the graph change? a. Increased slope b. Increased f o c. Both a and b 2. If 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) 1 2 3 5 10 15 f Lecture 9, p 5 Act 1  Solution 1. If Plancks constant were somewhat larger, but remained the same, how would the graph change? a. Increased slope b. Increased f o c. Both a and b 2. If 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) 1 2 3 5 10 15 f max ( ) stop KE e V h f f hf = = = From we can see that the slope increases, and f decreases. Lecture 9, p 6 Act 1  Solution 1. If Plancks constant were somewhat larger, but remained the same, how would the graph change? a. Increased slope b. Increased f o c. Both a and b 2. If 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) 1 2 3 5 10 15 f From we can see that the slope increases, and f decreases. The y intercept moves down, so the slope is unchanged and f increases.  max ( ) stop KE e V h f f hf = = = Lecture 9, p 7 Summary: Photon & Matter Waves Light (v = c) E = pc, so E = hc/ Slow Matter (v << c) KE = p 2 /2m, so KE = h 2 /2m 2 For electrons: 1240 eV nm photon E = 2 2 1.505 eV nm KE = Everything E = hf p = h/ Lecture 9, p 8 Act 2: Counting photons Which emits more photons , a 1mW cell phone (f = 830 MHz b = 0.36 m ) or a 1mW laser ( = 635 nm )? a) Laser emits more b) They emit the same number c) Cell phone emits more Lecture 9, p 9 Act 2: Solution Rate 5 9 0.36 m 5.7 10 635 10 m cell cell laser laser Rate Rate  = = = Because the cell frequency is much less than the optical frequency, each cellphone photon has much less energy. Therefore, you need many more of them to get the same total energy....
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This note was uploaded on 04/04/2011 for the course PHYSICS 214 taught by Professor Mestre during the Spring '11 term at University of Illinois at Urbana–Champaign.
 Spring '11
 MESTRE
 Magnetism, Energy, Work

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