hw7_solutions - PHY 251 Fall 2009: homework problem set 7,...

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Unformatted text preview: PHY 251 Fall 2009: homework problem set 7, due in the PHY 251 drop box in room A-129 by noon on Friday, Oct. 30. 1. Serway 6.15 Answer: (a) The potential energy from the Coulomb force is U = q 1 q 2 4 πǫ r . The q 1 q 2 terms are all going to have an electron charge times either -1 (electron) or +1 (proton), and the distances r are going to be 1 d , 2 d , or 3 d . We can then work out the potential seen by the particles from left to right, being careful not to double-count (that is, don’t count again the potential of a particle to the left of whatever one we’re now considering): Δ U 1 = e 2 4 πǫ d bracketleftBigg ( − 1)(+1) 1 + ( − 1)( − 1) 2 + ( − 1)(+1) 3 bracketrightBigg = e 2 4 πǫ d bracketleftbigg − 1 + 1 2 − 1 3 bracketrightbigg Δ U 2 = e 2 4 πǫ d bracketleftBigg (+1)( − 1) 1 + (+1)(+1) 2 bracketrightBigg = e 2 4 πǫ d bracketleftbigg − 1 + 1 2 bracketrightbigg Δ U 3 = − e 2 4 πǫ d bracketleftBigg ( − 1)(+1) 1 bracketrightBigg = e 2 4 πǫ d [ − 1] so that the net potential energy is U = Δ U 1 + Δ U 2 + Δ U 3 = e 2 4 πǫ d bracketleftbigg − 1 + 1 2 − 1 3 − 1 + 1 2 − 1 bracketrightbigg = e 2 4 πǫ d bracketleftbigg − 6 6 + 3 6 − 2 6 − 6 6 + 3 6 − 6 6 bracketrightbigg = e 2 4 πǫ d bracketleftbigg − 14 6 bracketrightbigg = − e 2 4 πǫ d 7 3 = − 7 e 2 12 πǫ d . (b) If we say that the electrons are restricted to a 1D box of width 3 d , that’s like saying L = 3 d in the infinite quantum well solution, and n = 1 for the lowest energy quantum state. Therefore the kinetic energy of allowed electron wavefunctions in the box is E k = E 1 = 1 2 π 2 ¯ h 2 2 m (3 d ) 2 = h 2 72 md 2 which should be doubled for two electrons. (c) The total energy is U + E k ; to find the value of d which minimizes the total energy we set the derivative with respect to d to be zero: ∂ ∂d ( U + E k ) = 0 = ∂ ∂d parenleftBigg − 7 e 2 12 πǫ d + h 2 36 md 2 parenrightBigg = parenleftBigg − ( − 1) 7 e 2 12 πǫ d 2 + ( − 2) h 2 36 md 3 parenrightBigg 7 e 2 12 πǫ d 2 = h 2 18 md 3 d = 4 πǫ h 2 42 me 2 = 4 π (8 . 85 × 10- 12 ) · (6 . 63 × 10- 34 ) 2 42 · (9 . 11 × 10- 31 ) · (1 . 602 × 10- 19 ) 2 = 4 . 98 × 10- 11 meters...
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This note was uploaded on 05/28/2011 for the course PHY 251 taught by Professor Rijssenbeek during the Fall '01 term at SUNY Stony Brook.

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hw7_solutions - PHY 251 Fall 2009: homework problem set 7,...

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