HOMEWORK #5 PROBLEM SET t DUE: Wednesday … to radiate 100 times the power P_sun radiation by the...
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HOMEWORK #5 PROBLEM SET DUE: Wednesday October 26 th 1. Calculate the temperature of a blackbody if the spectral distribution peaks at (a.) the record-breaking (detected from astronomical sources) gamma-ray wavelength of 1.24 x 10^-20 meters and at (b.) the Albany Oldies station (98.3 MHz). Hint: f = c / wavelength 2. Measurement of the wavelength at which the spectral distribution R(λ) from a certain star is maximum indicates that the star’s surface temperature is 3,000 K. If the star is also found to radiate 100 times the power P_sun radiation by the sun, then how big is this star? (As you have seen on earlier homework, the sun’s surface temperature is about 5,800 K.) 3. The threshold wavelength of potassium is 558 nm. What is the work function for potassium? 4. What is the stopping potential when light of wavelength 400 nm is used? 5. X-Rays from an old-school TV: The accelerating voltage of the electron in a typical late 1990s color television picture tube had been 25 kV. What was the minimum-wavelength x- ray produced when these electrons struck surfaces within the tube? 6. In a particular Compton scattering experiment, it is found that the incident wavelength λ1 is shifted by 1.5% when the scattering angle θ = 120°. (a.) What is the value of λ1? (b.) What will wavelength λ2 be for the shifted photon when the scattering angle instead is 75°? SOLUTIONS 1. (a.) lambda_max * T = 2.898 x 10 ^ -3 meter-Kelvin, so (1.24 x 10 ^ -20 m) * T = 2.898 x 10 ^ -3 m*K =>T=2.337 x 10^17 K (wow hot lots of °F, C!) (b.) wavelength = c / f = (299,792,458 m/s) / (98.3 x 10^6 Hz) = 3.04977 m T = 2.898 x 10 ^ -3 meter-Kelvin / 3.04977 m = 9.50 x 10^-4 K <~ 1 mK (cold!) 2.
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Transcript of HOMEWORK #5 PROBLEM SET t DUE: Wednesday … to radiate 100 times the power P_sun radiation by the...
HOMEWORK#5PROBLEMSET DUE:WednesdayOctober26th1.Calculatethetemperatureofablackbodyifthespectraldistributionpeaksat(a.)therecord-breaking(detectedfromastronomicalsources)gamma-raywavelengthof1.24x10^-20metersandat(b.)theAlbanyOldiesstation(98.3MHz).Hint:f=c/wavelength2.MeasurementofthewavelengthatwhichthespectraldistributionR(λ)fromacertainstarismaximumindicatesthatthestar’ssurfacetemperatureis3,000K.Ifthestarisalsofoundtoradiate100timesthepowerP_sunradiationbythesun,thenhowbigisthisstar?(Asyouhaveseenonearlierhomework,thesun’ssurfacetemperatureisabout5,800K.)3.Thethresholdwavelengthofpotassiumis558nm.Whatistheworkfunctionforpotassium?4.Whatisthestoppingpotentialwhenlightofwavelength400nmisused?5.X-Raysfromanold-schoolTV:Theacceleratingvoltageoftheelectroninatypicallate1990scolortelevisionpicturetubehadbeen25kV.Whatwastheminimum-wavelengthx-rayproducedwhentheseelectronsstrucksurfaceswithinthetube?6.InaparticularComptonscatteringexperiment,itisfoundthattheincidentwavelengthλ1isshiftedby1.5%whenthescatteringangleθ=120°.(a.)Whatisthevalueofλ1?(b.)Whatwillwavelengthλ2befortheshiftedphotonwhenthescatteringangleinsteadis75°?SOLUTIONS1.(a.)lambda_max*T=2.898x10^-3meter-Kelvin,so(1.24x10^-20m)*T=2.898x10^-3m*K=>T=2.337x10^17K(wowhotlotsof°F,C!)(b.)wavelength=c/f=(299,792,458m/s)/(98.3x10^6Hz)=3.04977mT=2.898x10^-3meter-Kelvin/3.04977m=9.50x10^-4K<~1mK(cold!)2.
3.e*V=KE_max=h*f–phiV=(h*f/e)–(phi/e)“threshold”èKE_max=0henceV=0phi/e=h*f/e=h*c/(e*lambda)=1,240eV-nm/558nm=2.22eV(substitutingforh*c)4.Thisisrelatedto#3(stillaboutpotassium).When400-nmlightisused,Visgivenby:V=h*c/(e*lambda)=phi/e=(1240eV*nm/400nm)–2.22eV=3.10eV–2.22eV=0.88eVNotethatthestoppingpotentialVinvoltsisnumericallyequaltotheKE_maxof(1/2*m*v^2)_maxineV.Thus,wehave:V=0.88V(olts)5.TheComptonwavelengthoflambda=h*c/Eistheminpossibleinthiscase.Lambda_min=1.24x10^3eV-nm/V=1.24x10^3eV-nm/25,000eV=0.050nm(because25kV=25,000V->25,000eVinenergy)6.
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