Lect09 - Lecture 9, p 1 Lecture 9, p 2 Lecture 9:...

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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 eV-nm/ 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 1-mW cell phone (f = 830 MHz b = 0.36 m ) or a 1-mW 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 cell-phone 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.

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Lect09 - Lecture 9, p 1 Lecture 9, p 2 Lecture 9:...

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