PROBLEM SET 1 - Dr. P. Lucas U of A MSE 110 LIGHT E = h υ...

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Dr. P. Lucas U of A MSE 110 LIGHT E = h υ υ =c/ λ h: Planck constant h = 6.626x10 -34 J.s Duality wave/particle of the light: two possible descriptions • Electromagnetic Energy: Light
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Dr. P. Lucas U of A MSE 110 Problem 1 An Infrared laser emits photons of wavelength λ = 785 nm. What is the energy of these photons in Joules? (5 pts) [Answer in J] λ = ν = c h h E Joules 10 53 . 2 m 10 785 sec m 10 3 s J 10 626 . 6 E 19 9 8 34 × = × = Ans: where c = 3 × 10 8 m /sec, and = 785 × 10 -9 m Therefore,
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Dr. P. Lucas U of A MSE 110 Problem 2 What is the energy of the infrared laser photons ( λ = 785 nm) in eV? eV 583 . 1 J 10 6 . 1 eV 1 J 10 53 . 2 E 19 19 = × × = Joules 10 53 . 2 m 10 785 sec m 10 3 s J 10 626 . 6 E 19 9 8 34 × = × = Ans: The definition of an eV is the energy gained by an electron accelerated by a potential of 1 V. Therefore it is equal to one V times the charge of the electron, E = 1 V × 1.6·10 -19 C. Hence 1 eV = 1.6·10 -19 Joules
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Dr. P. Lucas U of A MSE 110 Problem 3 The energy of an Argon ion laser photon is 2.41 eV. What is its frequency? (5 pts) [Answer in Hz (s -1 )] eV J s J eV h E 19 34 10 6 . 1 10 626 . 6 41 . 2 = = ν Hz 10 82 . 5 sec 10 82 . 5 14 1 14 × = × = ν Ans: E=h ν hence
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Dr. P. Lucas U of A MSE 110 Energy is only transferred in packets or quanta, hence only the photon with sufficient energy quanta can excite an electron. Energy is quantized. QUANTIZATION OF ENERGY Einstein’s photoelectric effect The blue photon posses enough energy to eject an electron form the surface and to provide kinetic energy to the electron. 2 2 1 v m h e + Φ = υ The electron is bound to the metal with energy Φ . It takes an energy Φ to extract an electron from the Na metal. Φ is the work function of the metal. Low frequency light (red) cannot transfer energy continuously to the Na atom and extract an electron.
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Dr. P. Lucas U of A MSE 110 Problem 4 Light of wavelength λ = 532 nm is used in a photoelectric experiment to irradiate a metal electrode. Which of the following metals should be used to obtain a current through the photoelectric tube? Metal 1: Φ =2.5eV Metal 2: Φ =1.8eV Metal 3: Φ =4.8eV Metal 4: Φ =7.1eV eV eV J m m s J E 33 . 2 10 6 . 1 10 532 sec 10 3 10 626 . 6 19 9 8 34 = × × × = λ = ν = c h h E Ans: The energy of the photon is Hence this photon will only be energetic enough to extract electrons from metal with work function Φ smaller the 2.33eV. Hence only metal 2 will show a current.
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Dr. P. Lucas U of A MSE 110 Problem 5 A UV laser of wavelength 182 nm is used to extract electron from a metallic cathode of work function Φ = 7.1eV. What is the maximum kinetic energy E k of the electron extracted from the cathode?
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This note was uploaded on 10/06/2009 for the course MSE 110 taught by Professor Lucas during the Spring '08 term at University of Arizona- Tucson.

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PROBLEM SET 1 - Dr. P. Lucas U of A MSE 110 LIGHT E = h υ...

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