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hw7_solutions

# hw7_solutions - PHY 251 Fall 2009 homework problem set 7...

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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 0 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 0 d bracketleftBigg ( 1)(+1) 1 + ( 1)( 1) 2 + ( 1)(+1) 3 bracketrightBigg = e 2 4 πǫ 0 d bracketleftbigg 1 + 1 2 1 3 bracketrightbigg Δ U 2 = e 2 4 πǫ 0 d bracketleftBigg (+1)( 1) 1 + (+1)(+1) 2 bracketrightBigg = e 2 4 πǫ 0 d bracketleftbigg 1 + 1 2 bracketrightbigg Δ U 3 = e 2 4 πǫ 0 d bracketleftBigg ( 1)(+1) 1 bracketrightBigg = e 2 4 πǫ 0 d [ 1] so that the net potential energy is U = Δ U 1 + Δ U 2 + Δ U 3 = e 2 4 πǫ 0 d bracketleftbigg 1 + 1 2 1 3 1 + 1 2 1 bracketrightbigg = e 2 4 πǫ 0 d bracketleftbigg 6 6 + 3 6 2 6 6 6 + 3 6 6 6 bracketrightbigg = e 2 4 πǫ 0 d bracketleftbigg 14 6 bracketrightbigg = e 2 4 πǫ 0 d 7 3 = 7 e 2 12 πǫ 0 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 0 d + h 2 36 md 2 parenrightBigg = parenleftBigg ( 1) 7 e 2 12 πǫ 0 d 2 + ( 2) h 2 36 md 3 parenrightBigg 7 e 2 12 πǫ 0 d 2 = h 2 18 md 3 d = 4 πǫ 0 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 = 0 . 0498 nm .

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