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Unformatted text preview: Solution Key for PSE#1 [1]+[2] January 29, 2007 1(a) P1.20] First we look at the energy of each photon, E photon = hÎ½ = hc Î» . Weâ€™re given the wavelength, Î» = 300 nm = 300 Â· 10 9 m . And so E photon = (6 . 626 Â· 10 34 J Â· s )(3 Â· 10 8 m Â· s 1 ) 300 Â· 10 9 m = 6 . 626 Â· 10 19 J Â· photon 1 Since 6 . 626 Â· 10 19 J is delivered by each photon, we can determine how many photons were absorbed since we are told that 3 . 25 Â· 10 3 J are absorbed total. # of photons = 3 . 25 Â· 10 3 J 6 . 626 Â· 10 19 J Â· photon 1 = 4 . 9 Â· 10 15 photons Using the equation for the photoelectric effect E photon = Î¦ + 1 2 m e v 2 e (1) We can calculate the kinetic energy seeing that we know the energy of each photon and work function ( 2 . 40 eV â†’ 3 . 84 Â· 10 19 J ). KE = E photon âˆ’ Î¦ = 6 . 626 Â· 10 19 J âˆ’ 3 . 84 Â· 10 19 J = 2 . 79 Â· 10 19 J Since KE = 1 2 m e v 2 e , then (remembering that 1 J = 1 kg Â· m 2 s 2 ) v e = radicalBig 2 KE m e = radicalBig 2 Â· 2 . 79 Â· 10 19 kg Â· m 2 s 2 9 . 11 Â· 10 31 kg = 7 . 82 Â· 10 5 m Â· s 1 1(b) P1.4] Simply use the de Broglie equation Î» = h p = h mv (2) Since weâ€™re told that the wavelength need to be on the order of a bond length, then Î» = 0 . 150 Â· 10 9 m , and so simply v = h mÎ» = 6 . 626 Â· 10 34 kg Â· m 2 Â· s 1 9 . 11 Â· 10 31 kg Â· . 150 Â· 10 9 m = 4 . 85 Â· 10 6 m Â· s 1 1(c) Bohr postulated that since matter has wavelike properties, if they are going to fit into an orbit, then nwavelengths must fit into a circular orbit (where n is an integer). The circumference of a circle is C = 2 Ï€r = nÎ» . Substituting equation 2, we get 2 Ï€r = n h mv , which can be rearranged to give 1 mvr = n h 2 Ï€ = n planckover2pi1 (3) Since angular momentum is L = mvr = n planckover2pi1 , we can see that the quantization of hydrogen comes from the simple postulate that nwavelengths fit into a circular orbit. 1(d) Since this is a homework problem Iâ€™m going to leave out most of the algebra, since thatâ€™s your job....so anyway, from the homework we started with equating two forces, the coulombic force, and the force exerted on a particle moving in a loop. In this case the electron (with charge e) is moving in a loop around the nucleus (with charge +Ze)...note this is not true but this is what Bohr postulated. F coulomb = Ze 2 4 Ï€Ç« ,F loop = m v 2 r Setting these two equations equal we get Ze 2 4 Ï€Ç« = m v 2 r (4) And combining with what Bohr postulated (what you derived in 1(d)), mvr = n planckover2pi1 , we effectively have two equations and two unknowns. Equation 3 can be substituted into equation 4, and with some simple algebra, can yield r n = Ç« n 2 h 2 Ï€Ze 2 m e (5) Equation 5 can be substituted back into equation 3, and rearranging, can yield...
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 Winter '07
 Lin
 pH, Quantum Chemistry, Energy, Schrodinger Equation, Fundamental physics concepts

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