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Unformatted text preview: 27. (a) We use the photoelectric eﬀect equation (Eq. 395) in the form hc/λ = Φ + Km . The work function depends only on the material and the condition of the surface, and not on the wavelength of the incident light. Let λ1 be the ﬁrst wavelength described and λ2 be the second. Let Km1 = 0.710 eV be the maximum kinetic energy of electrons ejected by light with the ﬁrst wavelength, and Km2 = 1.43 eV be the maximum kinetic energy of electrons ejected by light with the second wavelength. Then, hc hc = Φ + Km1 and = Φ + Km2 . λ1 λ2 The ﬁrst equation yields Φ = (hc/λ1 ) − Km1 . When this is used to substitute for Φ in the second equation, the result is (hc/λ2 ) = (hc/λ1 ) − Km1 + Km2 . The solution for λ2 is λ2 = = = hcλ1 hc + λ1 (Km2 − Km1 ) (1240 eV · nm)(491 nm) 1240 eV · nm + (491 nm)(1.43 eV − 0.710 eV) 382 nm . Here hc = 1240 eV · nm, calculated in Exercise 3, is used. (b) The ﬁrst equation displayed above yields Φ= hc 1240 eV · nm − Km1 = − 0.710 eV = 1.82 eV . λ1 491 nm ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Work, Light

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