A clean molybdenum surface offers sites that each hold at most one atom

PHYSICS · Multi-part · Fermi University Physics · Thermal, Statistical & Complex Systems · Thermodynamic potentials and the formal structure of thermodynamics · Statistical foundations and the equilibrium ensembles

Thermal Physics, Ch The grand canonical ensemble, ensemble equivalence and fluctuations

Problem

A clean molybdenum surface offers sites that each hold at most one atom, and it binds a helium atom and a krypton atom with the same energy: the occupied site has energy

-\varepsilon_0

with

\varepsilon_0 = 0.35\,\mathrm{eV}

, the empty site

0

, and the adsorbed atom has no other internal states. In one run the chamber holds helium (mass

4.00\,\mathrm{u}

) at

p = 100\,\mathrm{Pa}

and

T = 300\,\mathrm{K}

; in a second run it holds krypton (mass

83.8\,\mathrm{u}

) at the same

100\,\mathrm{Pa}

and

300\,\mathrm{K}

. Both gases are ideal and classical, and the sites do not interact with one another. A student argues: "the binding energy is the same and

p

and

T

are the same, so the coverage must be the same; if anything the krypton, being twenty-one times heavier, will stick better." Take

k_B = 1.381\times10^{-23}\,\mathrm{J\,K^{-1}}

h = 6.626\times10^{-34}\,\mathrm{J\,s}

1\,\mathrm{u} = 1.661\times10^{-27}\,\mathrm{kg}

1\,\mathrm{eV} = 1.602\times10^{-19}\,\mathrm{J}

. ((a)) Compute the coverage

\theta

in each run. ((b)) Which gas covers the surface more, and by what factor? Explain, in terms of the chemical potential and of the entropy the gas gives up per atom, why the mass enters at all, and why the binding energy alone cannot settle the question. ((c)) The surface carries

5.0\times10^{14}

equivalent independent sites. For the helium run, use this chapter's

N^{-1/2}

rule to estimate the relative wandering of the number of occupied sites, and say what that means for calling

\theta

a thermodynamic quantity.

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