 Solution

ID: 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

x = ν_{l} T

. Expanding

e^{- x} = 1 - x + x^{2} /2 - x^{3} /6 + \dots

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} ν_{l} T^{2} /2 + O (T^{3})

and the denominator contribution is

A_{l} ν_{l} T + O (T^{2})

For one mode, cancellation yields

T (T) = \frac{1}{ν} - \frac{T}{e ^{ν T} - 1}, T (T) \sim T /2 (T \to 0), T (T) \to 1/ ν (T \to \infty) .

The 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/ ν

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