PrtinBox

# PrtinBox - Particle in a Box 12 C=CC=CC=C | a | 10...

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Colby College Particle in a Box C=C–C=C–C=C | a | H Ψ = E Ψ h - 2 2m d 2 dx 2 Ψ + V(x) Ψ = E Ψ h - 2 2m d 2 dx 2 Ψ = E Ψ for 0 x a 0 2 4 6 8 10 12 0 2 4 6 8 10 E n /(h 2 /8ma ) x (Å) A sin kx : d Ψ dx = A k cos kx d 2 Ψ dx 2 = – A k 2 sin kx B cos kx : d Ψ dx = – B k sin kx d 2 Ψ dx 2 = – B k 2 cos kx LHS : RHS : h - 2 2m (– A k 2 sin kx ) = E A sin kx E = h - 2 k 2 2m k = (2mE) ½ h - General Solution: Ψ (x) = A sin kx + B cos kx Boundary Conditions: at x = 0: Ψ (x) = 0 cos(0) = 1 so B = 0 at x = a: Ψ (x) = 0 Ψ (a) = A sin ka = 0 so k = n π / a n = 1 kx = n π x a at a: ka = π n = 2 ka = 2 π E = n 2 h - 2 π 2 2ma 2 = n 2 h 2 8ma 2 Normalization - Ψ 2 dx = 1 x π /2 π 3 π /2 cos 0 = 1 sin 0 = 0 cos 2 π = 1 sin 2 π = 0 cos π /2 = 0 sin π /2 = 1 cos π = -1 sin π = 0 cos 3 π /2 = 0 sin 3 π /2 = -1

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Colby College = A 2 0 a sin 2 n π x a dx = 1 for particle in the box. let y = n
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## This document was uploaded on 02/22/2012.

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PrtinBox - Particle in a Box 12 C=CC=CC=C | a | 10...

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