826.3 The photoelectric effect When em radiation falls on the metal, electrons are emitted virtually instantaneously the time delay observed experimentally is about 10-s, regardless of the light intensity. If the em radiation behaves as a classical wave when the intensity of the light is low, it should take some time for enough energy to accumulate on a particular spot to liberate an electron .o For a given frequency of light, if we reverse the polarity of the potential difference and increase the value of the potential difference, the current in the circuit decreases to zero i e there is a stopping potential Vs 826.3 The photoelectric effect Ultraviolet 光电管 金属平板 1.oFcutolT 安培表 Frequency of incident light v(OHz v(1014Hz
11 When EM radiation falls on the metal, electrons are emitted virtually instantaneously; the time delay observed experimentally is about 10-9s, regardless of the light intensity. If the EM radiation behaves as a classical wave, when the intensity of the light is low, it should take some time for enough energy to accumulate on a particular spot to liberate an electron. For a given frequency of light, if we reverse the polarity of the potential difference and increase the value of the potential difference, the current in the circuit decreases to zero, i. e. there is a stopping potential Vs. §26.3 The photoelectric effect ν c ν ( ν §26.3 The photoelectric effect
826.3 The photoelectric effect 2. Einstein's photon theory Einstein imagined light to consist of wavelike particles, pockets(bundles)of electromagnetic energy, called photon. Ephoton =hv hv=l+KE ev Slope h KE my =ev KE max hv w=h V=-w+hy 826.3 The photoelectric effect Example 1: A low Helium-neon laser has a power output of 1.00 mw of light of wavelength 632, 8 nm.(a) Calculate the energy of each photon, expressing your result in joules and electron-volts;(b)Determine the number of photons emitted by the laser each second. Solution: (a)E=hhc (6.626×10-)(30×10°) 6328×10 =3.13×10-19J=1.96eV
12 hν = W + KE max Slope h eVs hν νc ν ν W KEmax -W 2. Einstein’s photon theory Ephoton = hν Einstein imagined light to consist of wavelike particles, pockets(bundles) of electromagnetic energy, called photon. s KE = mv = eV 2 max m 2 1 W = hν c eVs = −W + hν §26.3 The photoelectric effect Example 1: A low Helium-neon laser has a power output of 1.00 mW of light of wavelength 632,8 nm.(a) Calculate the energy of each photon, expressing your result in joules and electron –volts; (b) Determine the number of photons emitted by the laser each second. Solution: 3.13 10 J 1.96eV 632.8 10 (6.626 10 )(3.0 10 ) 19 9 34 8 = × = × × × = = = − − − λ ν hc (a) E h §26.3 The photoelectric effect
826.3 The photoelectric effect (b)N=1.00×10-3 =318×103 photons,/s 3.14×10 Example 2: Ultraviolet light of wavelength 200 nm is incident on a freshly polished iron surface. Find(a) the stopping potential; (b)the maximum kinetic energy of the liberated electrons; and (c) the speed of these fastest electrons lution:(a) hv=w+eV From Table 26.1 W=47eV=4.7×1.602×10-19=7.5×10-10J 826.3 The photoelectric effect hv-w hc/1-w then v =1.5V (b)The maximum kinetic energy hc KE =ev W=24×10-9J (c) The speed of fastest electrons mv2=eV=2.4×10-1J 2 2eV./2×2.4×10-9 v〓 m-V9.11×1031 =73×105m/s
13 3.18 10 photons/s 3.14 10 1.00 10 15 19 3 = × × × = − − (b) N Example 2: Ultraviolet light of wavelength 200 nm is incident on a freshly polished iron surface. Find (a) the stopping potential; (b)the maximum kinetic energy of the liberated electrons; and (c) the speed of these fastest electrons. Solution: (a) s hν = W + eV From Table 26.1 4.7eV 4.7 1.602 10 7.5 10 J −19 −19 W = = × × = × §26.3 The photoelectric effect then 1.5V / = − = − = e hc W e h W Vs ν λ (b) The maximum kinetic energy 2.4 10 J 19 max − = = −W = × hc KE eVs λ (c) The speed of fastest electrons 7.3 10 m/s 9.11 10 2 2 2.4 10 2.4 10 J 2 1 5 31 19 2 19 max = × × × × = = = = × − − − m eV v mv eV s s §26.3 The photoelectric effect