 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-2/a6/d/i/solution

Normalizing the geometric distribution requires

1 = θ_{0} \sum_{j = 0}^{\infty} x^{j} = \frac{θ _{0}}{1 - x}, θ_{0} = 1 - x, 0 \leq x < 1.

There are

M θ_{j}

sites containing

j

atoms each. Therefore the expected total population is

N_{ads} = M \sum_{j = 1}^{\infty} j θ_{j} = M (1 - x) \sum_{j = 1}^{\infty} j x^{j} = M (1 - x) \frac{x}{( 1 - x ) ^{2}} .

Consequently

N_{ads} = \frac{M x}{1 - x} = \frac{M K _{ads} p}{1 - K _{ads} p} .

The quantity counted is the total number of atoms in every layer. The number of nonempty columns is instead

M (1 - θ_{0}) = M x

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