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# Lect09 - Lecture 9 p 1 Lecture 9 Introduction to QM Review...

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Lecture 9, p 1

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Lecture 9, p 2 Lecture 9: Introduction to QM: Review and Examples S 1 S 2
Lecture 9, p 3 The work function: circle6 Φ is the minimum energy needed to strip an electron from the metal. circle6 Φ is defined as positive . circle6 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 f 0 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

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Lecture 9, p 4 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 f 0
Lecture 9, p 5 Act 1 - Solution 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 f 0 max 0 ( ) stop KE e V h f f hf = = - = - Φ From we can see that the slope increases, and f 0 decreases. - Φ

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Lecture 9, p 6 Act 1 - Solution 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 f 0 From we can see that the slope increases, and f 0 decreases. The y intercept moves down, so the slope is unchanged and f 0 increases. - Φ max 0 ( ) stop KE e V h f f hf = = - = - Φ
Lecture 9, p 7 Summary: Photon & Matter Waves 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 λ = Everything E = hf p = h/ λ

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Lecture 9, p 8 Act 2: Counting photons Which emits more photons , a 1-mW cell phone (f = 830 MHz barb2right λ = 0.36 m ) or a 1-mW laser ( λ = 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 ∝ λ 5 9 0.36 m 5.7 10 635 10 m cell cell laser laser Rate Rate λ λ - = = = × × 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. Cell phones actually emit ~1W barb2right ~10 24 photons/sec Which emits more photons , a 1-mW cell phone (f = 830 MHz barb2right λ = 0.36 m ) or a 1-mW laser ( λ = 635 nm )?

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Lect09 - Lecture 9 p 1 Lecture 9 Introduction to QM Review...

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