# ch38 - Chapter 38 1. (a) With E = hc/min = 1240 eVnm/min =...

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1455Chapter 381. (a) WithE=hc/min= 1240 eV·nm/min= 0.6 eV, we obtain= 2.1103nm = 2.1m.(b) It is in the infrared region.2. Let122m vEhcephotonand solve forv:v2hcme2hcmec2c2c2hcmec22.998108m/s2 1240eVnm588nm511103eV8.6105m/s.Sincevc,the nonrelativistic formulaKmv122may be used. Themec2value ofTable 37-3 and1240eV nmhcare used in our calculation.3. LetRbe the rate of photon emission (number of photons emitted per unit time) of theSun and letEbe the energy of a single photon. Then the power output of the Sun is givenbyP = RE. NowE = hf = hc/,whereh= 6.62610–34J·s is the Planck constant,fis the frequency of the light emitted,andis the wavelength. ThusP = Rhc/and2645348550nm3.9 10W1.0 10photons/s.6.63 10J s2.998 10 m/sPRhc4. We denote the diameter of the laser beam asd. The cross-sectional area of the beam isA =d2/4. From the formula obtained in Problem 38-3, the rate is given by
CHAPTER 381456RAPhcd2/ 44 633nm5.0103W6.631034Js2.998108m/s3.0103m22.31021photons/m2s .5. The energy of a photon is given byE = hf, wherehis the Planck constant andfis thefrequency. The wavelengthis related to the frequency byf = c, soE = hc/. Sinceh=6.62610–34J·s andc= 2.998108m/s,hc6 626102 99810160210101240348199...J sm / sJ / eVm / nmeV nm.chchchchThus,E1240eV nm.With= (1, 650, 763.73)–1m = 6.057802110–7m = 605.78021 nm,we find the energy to beEhc1240605780212 047eV nmnmeV...6. The energy of a photon is given byE = hf, wherehis the Planck constant andfis thefrequency. The wavelengthis related to the frequency byf = c, soE = hc/. Sinceh=6.62610–34J·s andc= 2.998108m/s,hc6 626102 99810160210101240348199...J sm / sJ / eVm / nmeV nm.chchchchThus,E1240eV nm.With589 nm, we obtain1240eV nm2.11eV.589nmhcE7. The rate at which photons are absorbed by the detector is related to the rate of photonemission by the light source viaabsabsemit2(0.80).4ARRrGiven that62abs2.0010mAand3.00 m,rwithabs4.000 photons/s,Rwe find therate at which photons are emitted to be
1457228emitabs62abs44(3.00 m)4.000 photons/s2.83 10 photons/s(0.80)(0.80)(2.0010m )rRRA.Since the energy of each emitted photon isph1240 eV nm2.48 eV500nmhcE,the power output of source is8810emitemitph2.83 10 photons/s (2.48 eV)7.010 eV/s1.1 10W.PRE8. The rate at which photons are emitted from the argon laser source is given byR =P/Eph, whereP= 1.5 W is the power of the laser beam andEph=hc/is the energy of

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Compton scattering, ev
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