p39_023 - V is given by K m = eV , so V = K m /e = (2 . 00...

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23. (a) The kinetic energy K m of the fastest electron emitted is given by K m = hf Φ=( hc/λ ) Φ, where Φ is the work function of aluminum, f is the frequency of the incident radiation, and λ is its wavelength. The relationship f = c/λ was used to obtain the second form. Thus, K m = 1240 eV · nm 200 nm 4 . 20 eV = 2 . 00 eV where the result of Exercise 3 is used. (b) The slowest electron just breaks free of the surface and so has zero kinetic energy. (c) The stopping potential
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Unformatted text preview: V is given by K m = eV , so V = K m /e = (2 . 00 eV) /e = 2 . 00 V. (d) The value of the cutoF wavelength is such that K m = 0. Thus hc/ = or = hc/ = (1240 eV nm) / (4 . 2 eV) = 295 nm. If the wavelength is longer, the photon energy is less and a photon does not have sucient energy to knock even the most energetic electron out of the alu-minum sample....
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