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Unformatted text preview: 3810 Chapter 38 Its root is 4.965, so 4.965.(4.965)hckT==(b)3483m23(6.63 10J s)(3.00 10 m s)2.90 10 m K.(4.965)(4.965)(1.38 10J K)hcTk===38.48. IDENTIFY: Since the stars radiate as blackbodies, they obey the StefanBoltzmann law. SET UP: The StefanBoltzmann law says that the intensity of the radiation is I= T 4, so the total radiated power is P = AT 4. EXECUTE: (a)I = T 4= (5.67 10824W/mK)(24,000 K)4= 1.9 1010W/m2(b)Wiens law gives m= (0.00290 m K)/(24,000 K) = 1.2 107m = 120 nm This is not visible since the wavelength is less than 400 nm. (c)P = AI 4R2= P/I = (1.00 1025W)/(1.9 1010W/m2) which gives RSirius= 6.51 106m = 6510 km. RSirius/Rsun= (6.51 106m)/(6.96 109m) = 0.0093, which gives RSirius= 0.0093 Rsun1% Rsun(d)Using the StefanBoltzmann law, we have 24424sunsun sunsun sunsunsun424SiriusSirius SiriusSirius SiriusSiriusSirius44PATRTRTPATRTRT===24sunsunSiriussun5800 K390.0093524,000 KPRPR == EVALUATE: Even though the absolute surface temperature of Sirius B is about 4 times that of our sun, it radiates only 1/39 times as much energy per second as our sun because it is so small. 38.49. Eq. (38.32): 2252( )but11for(1)2xhckThcxIexxe==++++"254221()Eq.()hcckTxIhc kT==V(38.31), whichis Rayleighs distribution. 38.50. (a)Wiens law: mkT=. 38m2.90 10K m9.7 10 m97 nm30,000 K===This peak is in the ultraviolet region, which is notvisible. The star is blue because the largest part of the visible light radiated is in the blue violet part of the visible spectrum (b)4PAT=(StefanBoltzmann law) 26824249W(100, 000)(3.86 10W)5.67 10(4)(30,000 K)m K8.2 10 mRR==9starsun88.2 10 m126.96 10 mRR==(c) The visual luminosity is proportional to the power radiated at visible wavelengths. Much of the power is radiated nonvisible wavelengths, which does not contribute to the visible luminosity. 38.51. IDENTIFYand SET UP: Use cf=to relate frequency and wavelength and use Ehf=to relate photon energy and frequency. EXECUTE: (a) One photon dissociates one AgBr molecule, so we need to find the energy required to dissociate a single molecule. The problem states that it requires 51.00 10 Jto dissociate one mole of AgBr, and one mole contains Avogadros number 23(6.02 10 )of molecules, so the energy required to dissociate one AgBr is 519231.00 10 J/mol1.66 10J/molecule.6.02 10 molecules/mol=The photon is to have this energy, so 19191.66 10J(1eV/1.602 10J)1.04 eV.E==(b) hcE=so 348619(6.626 10J s)(2.998 10 m/s)1.20 10 m1200 nm1.66 10JhcE====(c) cf=so 81462.998 10 m/s2.50 10 Hz1.20 10 mcf===(d) 34626(6.626 10J s)(100 10 Hz)6.63 10JEhf===261976.63 10J(1 eV/1.602 10J)4.14 10 eVE==Photons, Electrons, and Atoms 3811(e) EVALUATE: A photon with frequency...
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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.
 Spring '06
 Buchler
 Physics

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