PracticeExam1solution-Q1Q2 - Solution Key for PSE#1[1[2 1(a...

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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 = = hc λ . We’re given the wavelength, λ = 300 nm = 300 · 10 - 9 m . And so E photon = (6 . 626 · 10 - 34 J · s )(3 · 10 8 m · s - 1 ) 300 · 10 - 9 m = 6 . 626 · 10 - 19 J · photon - 1 Since 6 . 626 · 10 - 19 J is delivered by each photon, we can determine how many photons were absorbed since we are told that 3 . 25 · 10 - 3 J are absorbed total. # of photons = 3 . 25 · 10 - 3 J 6 . 626 · 10 - 19 J · photon - 1 = 4 . 9 · 10 15 photons Using the equation for the photoelectric effect E photon = Φ + 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 3 . 84 · 10 - 19 J ). KE = E photon Φ = 6 . 626 · 10 - 19 J 3 . 84 · 10 - 19 J = 2 . 79 · 10 - 19 J Since KE = 1 2 m e v 2 e , then (remembering that 1 J = 1 kg · m 2 s - 2 ) v e = radicalBig 2 KE m e = radicalBig 2 · 2 . 79 · 10 - 19 kg · m 2 s - 2 9 . 11 · 10 - 31 kg = 7 . 82 · 10 5 m · s - 1 1(b) P1.4] Simply use the de Broglie equation λ = h p = h mv (2) Since we’re told that the wavelength need to be on the order of a bond length, then λ = 0 . 150 · 10 - 9 m , and so simply v = h = 6 . 626 · 10 - 34 kg · m 2 · s - 1 9 . 11 · 10 - 31 kg · 0 . 150 · 10 - 9 m = 4 . 85 · 10 6 m · s - 1 1(c) Bohr postulated that since matter has wavelike properties, if they are going to fit into an orbit, then n-wavelengths must fit into a circular orbit (where n is an integer). The circumference of a circle is C = 2 πr = . Substituting equation 2, we get 2 πr = n h mv , which can be rearranged to give 1
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mvr = n h 2 π = 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 n-wavelengths fit into a circular orbit. 1(d) Since this is a homework problem I’m going to leave out most of the algebra, since that’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 πǫ 0 ,F loop = m v 2 r Setting these two equations equal we get Ze 2 4 πǫ 0 = 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 = ǫ 0 n 2 h 2 πZe 2 m e (5) Equation 5 can be substituted back into equation 3, and rearranging, can yield v n = Ze 2 2 ǫ 0 nh (6) Now we know that the total energy of a system is E total = E kinetic + E potential . The potential energy for a charged
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