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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 Schrodinger Equation in the two regions: h 2 2 m 2 x 2 ( x ) = E ( x ) , x < h 2 2 m 2 x 2 + V ( x ) = E ( x ) , x (10.2) or 2 x 2 ( x ) + k 2 ( x ) = 0 , k = 2 mE h x < 2 x 2 ( x ) + k 2 ( x ) = 0 , k = radicalBig 2 m ( E V ) h x (10.3) or 2 x 2 ( x ) = k 2 ( x ) , x < 2 x 2 ( x ) = k 2 ( 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 ( x ) = exp { sx } , which can be confirmed from the fact that then 2 x 2 ( x ) = 2 x 2 exp { sx } = s 2 exp { sx } which also yields a constant times the function. Thus, upon substituting ( 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 = k x < (10.6) s = k x (10.7) and the solutions in the two regions are: I ( x ) = A exp { k x } + B exp { k x } (10.8) II ( x ) = C exp { kx } + D exp { kx } (10.9) 4. Now the boundary conditions . (a) Continuity of the wavefunction at x =0: I ( x = 0) = 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 I ( x = 0) = II ( x = 0) k ( A B ) = k ( 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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 Winter '11
 AmarFlood

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