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