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Unformatted text preview: Physical Chemistry Course Number: C362 10 A onedimensional collision problem We will now treat a onedimensional collision problem. 1. Consider a step potential: V ( x ) = 0 x < V ( x ) = V x ≥ (10.1) So this is an unbound system. But what does this remind us of? A half chemical reaction, but where the potential has a straight edges? See figure on the beard. 2. The timeindependent Schr¨odinger Equation in the two regions: − ¯ h 2 2 m ∂ 2 ∂x 2 ψ ( x ) = Eψ ( x ) , x < − ¯ h 2 2 m ∂ 2 ∂x 2 + V ψ ( x ) = Eψ ( x ) , x ≥ (10.2) or ∂ 2 ∂x 2 ψ ( x ) + k 2 ψ ( x ) = 0 , k = √ 2 mE ¯ h x < ∂ 2 ∂x 2 ψ ( x ) + k 2 ψ ( x ) = 0 , k = radicalBig 2 m ( E − V ) ¯ h x ≥ (10.3) or ∂ 2 ∂x 2 ψ ( x ) = − k 2 ψ ( x ) , x < ∂ 2 ∂x 2 ψ ( x ) = − k 2 ψ ( x ) . x ≥ (10.4) Chemistry, Indiana University 49 c circlecopyrt 2011, Srinivasan S. Iyengar (instructor) Physical Chemistry Course Number: C362 3. And again, as we have seen a number of times before, these equations are essentially just a function differentiated twice (left side of the above equa tions), yields the same function back times a constant (right side of the above equations). Hence the solution in general is ψ ( x ) = exp { sx } , which can be confirmed from the fact that then ∂ 2 ∂x 2 ψ ( x ) = ∂ 2 ∂x 2 exp { sx } = s 2 exp { sx } which also yields a constant times the function. Thus, upon substituting ψ ( x ) = exp { sx } in the equations above, we obtain s 2 + k 2 = 0 , k = √ 2 mE ¯ h x < s 2 + k 2 = 0 , k = radicalBig 2 m ( E − V ) ¯ h x ≥ (10.5) and s = ± ık x < (10.6) s = ± ık x ≥ (10.7) and the solutions in the two regions are: ψ I ( x ) = A exp { ık x } + B exp {− ık x } (10.8) ψ II ( x ) = C exp { ıkx } + D exp {− ıkx } (10.9) 4. Now the boundary conditions . (a) Continuity of the wavefunction at x =0: ψ I ( x = 0) = ψ II ( x = 0) A + B = C + D (10.10) (b) We require that the derivative of the wavefunction be continuous at x =0. Note: We never used this in the particleinabox. Why are we using this here? Homework ψ ′ I ( x = 0) = ψ ′ II ( x = 0) ık ( A − B ) = ık ( C − D ) (10.11) Chemistry, Indiana University 50 c circlecopyrt 2011, Srinivasan S. Iyengar (instructor) Physical Chemistry Course Number: C362 (c) Equations ( 10.10 ) and ( 10.11 ) can be used to eliminate B : C = 2 k k + k A − k − k k + k D (10.12) which is obtained by multiplying Eq. ( 10.10 ) by k and adding with Eq. ( 10.11 ). We can also use Equations ( 10.10 ) and ( 10.11 ) to eliminate C : B = k − k k + k A + 2 k k + k D (10.13) which is obtained by multiplying Eq. ( 10.10 ) by k and subtracting from Eq. ( 10.11 )....
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This note was uploaded on 01/17/2012 for the course C 362 taught by Professor Amarflood during the Winter '11 term at Indiana.
 Winter '11
 AmarFlood

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