Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 38 | c_ X108 m s 577 x 10 Hz 1 x 10 7 m h P 1 x 10 34 J s x 10 7 m x 10 27 kg ms E pc x 10 2 7 kg ms x 108 ms x 10 9 J eV. a Pt W x 10 3s J x 1016eV. b hf x 10 9J eV. 1 Pt c x 1016. hf a E hf f E W 10-19 JeV x h h x 10 31 J s b 108 ms f x 1020 Hz x 10 3 m. 1 c 1 is of the same magnitude as a nuclear radius. 384 dN dE dt P P1 W x 10 7 m dt dE dN hf hc hc x 1019 photons sec. 12 . mv hf - 2 max I x 10 -34J s 7 L eV x 10-19 J eV x 10 -7m v x 10-20 J 2 10-20 J max J------------------ x 105 ms. x 10-31 kg E hf - hf- h x 1015Hz---------- 10 x 10-7m J eV. a f c k x 1014 Hz b Each photon has energy E hf x 10 J. Source emits 75 J s so 75 J s x 10 photons s x 1020 photons s c No they are different. The frequency depends on the energy of each photon and the number of photons per second depends on the power output of the source. For red light k 700 nm 0 h h k x 10 34 J s x 108 ms 700 x 10-9 m x 10-19 J 1eV x 10 9 J J eV a For a particle with mass K p2 2m. p2 2p1 means K2 4K1. b For a photon E pc. p2 2p1 means E2 2E1. Kmax hf -0 max Use the information given for k 400 nm to find 0 0 hf-m. 6 626x 10-34 J 2 998x 1 - eV 10-19 max 400 x 10-9 m x 10-19J Now calculate K for k 300 nm max hf - 6-626 x 10-34 J 2 998 x ms 10-19 j max 300 x 10-9 m x 10-19 J eV a The work function 0 hf - eV - eV0 1 0 x 10-34j s 3 00 x 108ms - x 10 -19C x 10-7m x 10-19J. . . h hc The threshold frequency implies 0 hfth 1 h 1 th 0 10-34 1ms 10- m. x 10J b 0 x 10-19 J eV as found in part a and this is the value from Table . V V. a From Eq. V 1 fhc x 10-15 eV s x 108 m s e 1 0J x 10-7 m b The
