This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY 251 Fall 2009: homework problem set 7, due in the PHY 251 drop box in room A129 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 doublecount (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...
View
Full
Document
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.
 Fall '01
 Rijssenbeek
 Physics, Energy, Force, Potential Energy, Work

Click to edit the document details