Problem $2(50$ points$)$

Consider a reaction between two particles that interact through the following potential energy function,

$V(r)=-\frac{d^6{E}_d}{r^6}$ for $r>d$

$V(r)=$ for $rd$

where $E_d$ is the well depth, $r$ is the interparticle separation and $d$ is the hard sphere collision diameter. The centrifugal barrier $V_{\text{e}\text{f}\text{f}}^{**}$ is given by the equation:

$V_{\text{e}\text{f}\text{f}}^{**}=C(\frac{b}{d}{)}^3(\frac{E^3}{E_d}{)}^{\frac{1}{2}}$

where $C=0\mathrm{.}385($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 $(b_{\text{m}\text{a}\text{x}})$ 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 $(E_{\text{c}\text{r}\text{i}\text{t}}),$ where $E_{\text{c}\text{r}\text{i}\text{t}}$ is a constant and $E_{\text{c}\text{r}\text{i}\text{t}}>E_d.$ Assume that the potential energy is equal to $)=(d$ at close contact. Write an expression for the radial kinetic energy $(T_r)$ at close contact. Does $T_r$ 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.

${}_R=d^2(\frac{E-E^{**}}{E})$

Part d$.$ Calculate the fraction of close contact collisions that result in reaction when $E=2E_d,$ given that $E_{\text{c}\text{r}\text{i}\text{t}}=1\mathrm{.}5E_d.$

$3$