Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section E / 27 / c / Solution

Set $x={\nu }_{l}T$. Expanding $e^{-x}=1-x+x^2/2-x^3/6+\cdots$ gives $1-e^{-x}=x+O(x^2)$ and $1-e^{-x}(1+x)=x^2/2+O(x^3)$. The numerator contribution is therefore $A_l{\nu }_l{T}^2/2+O(T^3)$ and the denominator contribution is $A_l{\nu }_{l}T+O(T^2)$.

For one mode, cancellation yieldsThe curve rises monotonically from zero, initially with slope $1/2$, and rounds towards a horizontal plateau. On logarithmic axes its short-time slope is one. The crossover occurs at observation times of order $1/\nu$.