1DScatt

1DScatt - Physical Chemistry Course Number C362 10 A...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physical Chemistry Course Number: C362 10 A one-dimensional collision problem We will now treat a one-dimensional 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 time-independent 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 particle-in-a-box. 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 )....
View Full Document

This note was uploaded on 01/17/2012 for the course C 362 taught by Professor Amarflood during the Winter '11 term at Indiana.

Page1 / 11

1DScatt - Physical Chemistry Course Number C362 10 A...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online