ps01soln

# ps01soln - 1 Physics 316 Cornell University Solution for...

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1 Physics 316 Solution for homework 1 Spring 2005 Cornell University Steve Drasco I. EXERCISE 1 Express Planck’s radiation formula for the emitted power per area R T ( ν ) = 2 πh c 2 ν 3 e kT - 1 (1.1) in terms of the wavelength λ = c/ν . The result should have the form ˜ R T ( λ ) = F ( λ ) dλ , (1.2) for which you have to determine F ( λ ) . Consider radiation in an arbitrary frequency-interval I defined by ν 1 ν ν 2 . The corresponding wavelength- interval is given by λ 2 λ λ 1 , where we have defined λ 1 = c/ν 1 and λ 2 = c/ν 2 . Let the P I be the power per unit area radiated in this interval. We have learned that this quantity can be expressed as P I = Z ν 2 ν 1 dν R T ( ν ) = Z ν 2 ν 1 2 πh c 2 ν 3 e kT - 1 · - 1 . (1.3) We are asked to express this quantity as an integral over wavelength P I = Z λ 1 λ 2 dλ F T ( λ ) , (1.4) for some function F T ( λ ), which we must determine. We do so by changing the integration variable in the frequency integral from ν to λ = c/ν . The differentials for these two variables are related by = = - c ν 2 = - λ 2 c dν. (1.5) Performing this operation gives P I = Z c/ν 2 c/ν 1 - c λ 2 · R T ( ν = c/λ ) = Z λ 1 λ 2 c λ 2 2 πh c 2 c λ · 3 e hc λkT - 1 · - 1 = Z λ 1 λ 2 2 πhc 2 λ - 5 e hc λkT - 1 · - 1 . (1.6) This means that the function F T ( λ ) must be F T ( λ ) = 2 πhc 2 λ - 5 e hc λkT - 1 · - 1 . (1.7) Another equation which equivalently states our result is R T ( ν ) = - F T ( λ ) . II. EXERCISE 2 A. A 1kW radio transmitter operates at a frequency of 900kHz. How many photons per second does it emit. Use Planck’s formula for the energy per photon, E = . An ideal radio transmitter radiates at a rate of 1 kW. Since the transmitter is tuned to a fixed frequency ( ν = 900 kHz) each photon which it emits has an energy of = (6 . 626 × 10 - 34 Js)(9 × 10 5 Hz) = 5 . 963 × 10 - 28 J. The rate at which this transmitter emits photons is then given by (10 3 W) / (5 . 963 × 10 - 28 J) = 1 . 677 × 10 30 Hz. 2 B. Approximate the emission of a 100W light bulb as that of a black body at 4000K. How many photons are emitted per second? You may use R 0 x 3 / ( e x - 1) dx = π 4 / 15 and R 0 x 2 / ( e x - 1) dx = 2 ζ (3) = 2 . 4 . We now derive the rate at which photons are emitted by a 100W light bulb under the assumption that the bulb’s filament is a black body with temperature T = 4000 K. The power per unit radiating area for photons with frequencies ranging from ν to ν + is given by R T ( ν ) . Since each of these photons has energy , the number flux

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