ch38 tủ tài liệu training

1 (a) Let E = 1240 eV·nm/λmin = 0.6 eV to get λ = 2.1 × 103 nm = 2.1 µm (b) It is in the infrared region The energy of a photon is given by E = hf, where h is the Planck constant and f is the frequency The wavelength λ is related to the frequency by λf = c, so E = hc/λ Since h = 6.626 × 10–34 J·s and c = 2.998 × 108 m/s, c6.626 × 10 hc = × 10 c1602 −34 hc h = 1240 eV ⋅ nm J ⋅ s 2.998 × 108 m / s −19 hc h J / eV 10−9 m / nm Thus, E= 1240eV ⋅ nm λ With λ = 589 nm , we obtain E= hc 1240eV ⋅ nm = = 2.11eV λ 589nm Let R be the rate of photon emission (number of photons emitted per unit time) of the Sun and let E be the energy of a single photon Then the power output of the Sun is given by P = RE Now E = hf = hc/λ, where h is the Planck constant, f is the frequency of the light emitted, and λ is the wavelength Thus P = Rhc/λ and b gc h 550 nm 3.9 × 1026 W λP × 1045 photons / s R= = = 10 −34 hc 6.63 × 10 J ⋅ s 2.998 × 10 m / s c hc h We denote the diameter of the laser beam as d The cross-sectional area of the beam is A = πd 2/4 From the formula obtained in problem 3, the rate is given by ( 633nm ) ( 5.0 ×10−3 W ) R λP = = A hc ( πd / ) π ( 6.63 ×10−34 J ⋅ s )( 2.998 ×108 m/s )( 3.5 ×10−3 m )2 = 1.7 ×1021 photons m2 ⋅ s Since λ = (1, 650, 763.73)–1 m = 6.0578021 × 10–7 m = 605.78021 nm, the energy is (using the fact that hc = 1240eV ⋅ nm ), E= hc 1240 eV ⋅ nm = = 2.047 eV λ 605.78021 nm Let hc me v = E photon = λ and solve for v: 2hc = λme v= c 2hc 2hc c =c λme c λ me c c h b1240 eV ⋅ nmg = 8.6 × 10 m / s h b5902nm gc511 × 10 eVh = 2.998 × 108 m / s Since v