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Problem 2 ( 5 0 points ) Consider a reaction between two particles that interact through the following potential energy function, V ( r )

Problem 2(50 points)
Consider a reaction between two particles that interact through the following potential energy function,
V(r)=-d6Edr6 for r>d
V(r)= for rd
where Ed is the well depth, r is the interparticle separation and d is the hard sphere collision diameter. The centrifugal barrier Veff** is given by the equation:
Veff**=C(bd)3(E3Ed)12
where C=0.385(positive constant),E is the initial energy and b is the impact parameter.
Part a. Derive an equation for the cross section for a "close contact" collision (CC), which is defined as a collision in which the particles approach to a distance r=d. Express the equation for the cross section in terms of the hard sphere collision diameter, the well depth and the initial energy E of the colliding pair. Hint: For a given E, only collisions with impact parameters less than a certain value ) can overcome centrifugal repulsion and make close contact.
Part b. Reaction occurs for collisions that make close contact and only when the radial kinetic energy is greater than a critical energy (Ecrit), where Ecrit is a constant and Ecrit>Ed. Assume that the potential energy is equal to )=(d at close contact. Write an expression for the radial kinetic energy (Tr) at close contact. Does Tr at close contact increase or decrease with b?
Part c. For the conditions given in Part b, show that the reaction cross section (R) is given by the expression below. Write the constant E** in terms of other constants given in this problem.
R=d2(E-E**E)
Part d. Calculate the fraction of close contact collisions that result in reaction when E=2Ed, given that Ecrit=1.5Ed.
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