1071_PartUniversity Physics Solution

1071_PartUniversity Physics Solution - Photons, Electrons,...

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Photons, Electrons, and Atoms 38-15 38.67. IDENTIFY: Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the Stefan- Boltzmann law apply to its radiation. SET UP: Wien’s displacement law is 3 peak 2.90 10 m K T λ ×⋅ = , and the Stefan-Boltzmann law says that the intensity of the radiation is I = σ T 4 , so the total radiated power is P = AT 4 . EXECUTE: (a) First use Wien’s law to find the peak wavelength: m = (2.90 × 10 –3 mK )/(3000 K) = 9.667 × 10 –7 m Call N the number osf photons/second radiated. N × (energy per photon) = IA = AT 4 . N ( hc/ m ) = AT 4 . 4 m AT N hc λσ = . 78 2 4 8 2 4 34 8 (9.667 10 m)(5.67 10 W/m K )(4 )(600 6.96 10 m) (3000 K) (6.626 10 J s)(3.00 10 m/s) N π −− ×× ⋅× × = ×⋅ × . N = 5 × 10 49 photons/s. (b) 2 4 42 4 B B BB S 4 SS S 4 600 3000 K 4 5800 K IA AT RT R R σπ ⎛⎞ == = ⎜⎟ ⎝⎠ = 3 × 10 4 EVALUATE: Betelgeuse radiates 30,000 times as much energy per second as does our sun! 38.68. IDENTIFY: The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody. The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength radiated, but a thermophile in the water measures the wavelength and frequency of the light in the water. SET UP: By the Stefan-Boltzman law, the net power radiated by the blackbody is () 44 sphere water dQ T dt =− . Since this heat evaporates water, the rate at which water evaporates is v dQ dm L dt dt = . Wien’s displacement law is 3 m 2.90 10 m K T = , and the wavelength in the water is w = 0 / n . EXECUTE: (a) The net radiated heat is sphere water dQ T dt and the evaporation rate is v dQ dm L dt dt = , where dm is the mass of water that evaporates in time dt . Equating these two rates gives v sphere water dm LA T T dt . ( ) 24 4 sphere water v 4 T dm dt L = . 82 4 2 4 4 4 3 5.67 10 W/m K 4 (0.120 m) (498 K) (373 K) 1.92 10 kg/s 0.193 g/s 2256 10 J/Kg dm dt ⎡⎤ ⎣⎦ × = × (b) (i) Wien’s law gives m = (0.00290 m K )/(498 K) = 5.82 × 10 –6 m But this would be the wavelength in vacuum. In the water the thermophile organism would measure w = 0 / n = (5.82 × 10 –6 m)/1.333 = 4.37 × 10 –6 m = 4.37 μm (ii) The frequency is the same as if the wave were in air, so f = c/ 0 = (3.00 × 10 8 m/s)/(5.82 × 10 –6 m) = 5.15 × 10 13 Hz EVALUATE: An alternative way is to use the quantities in the water: 0 / / cn f n = = c/ 0 , which gives the same answer for the frequency. An organism in the water would measure the light coming to it through the water, so the wavelength it would measure would be reduced by a factor of 1/ n . 38.69. IDENTIFY: The energy of the peak-intensity photons must be equal to the energy difference between the n = 1 and the n = 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must be for it to radiate with its peak intensity at this wavelength. SET UP: In the Bohr model, the energy of an electron in shell n is 2 13.6 eV n E n , and Wien’s displacement law is 3 m 2.90 10 m K T = . The energy of a photon is E = hf = hc/ .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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1071_PartUniversity Physics Solution - Photons, Electrons,...

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