hw8_solution

hw8_solution - Physics 3C Webassign Homework 8 Solutions:...

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Physics 3C Webassign Homework 8 Solutions: Practice using E = hf and λ =c/f. e.g. E = (6.626e-34 J s)(527e12 1/s)(1eV/1.6e-19 J) = 2.18 eV λ = (3.e8m/s)/(527e12 Hz) = 5.69e-7m = 569nm, this is visible light Each photon has an energy E = hf = (6.626e-34 J s)(96.0e6 Hz) = 6.36e-26 J P = (120e3 J/s) => (120e3 J/s)/(6.36e-26 J/photon) = 1.89e30 photons/s a) e Δ V = hc/ λ - φ , which implies φ = (1240 nm eV)/(546.1nm) -0.882 eV = 1.39eV b) e Δ V = hc/ λ - φ = (1240 nm eV)/(629.0 nm) – 1.39eV = 0.583V
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a) kinetic energy of electron: K = 0.5 m v^2 = 0.5(9.11e-31 kg)(1.25e6 m/s)^2 = 7.117e-19 J = 4.44eV Initial energy of the photon E = hc/ λ = 1240 eV nm/(0.790nm) = 1.5696e3 eV Final energy of photon Ef = E – K = 1.5652e3eV λ f = hc/Ef = 1240eV nm/(1.565e3 eV) = 0.79225nm λ f - λ = 2.25pm. b) λ f - λ = λ c(1-cos θ ) => cos θ = 1 - (2.25pm)/(2.43pm) = 0.074 => 85.7 degrees E_0
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λ = h/p = h/mv = (6.626e-34 J s)/(1.67e-27 kg)/(1.e6 m/s) = 3.61e-13m For the first bright wavelength, m=1
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hw8_solution - Physics 3C Webassign Homework 8 Solutions:...

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