Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 64 / b

The coherent neutron scattering length density is atomic number density times scattering length:The density was converted to $\mathrm{g}{\mathrm{m}}^{-3}$ to match the molar mass in grams per mole, and $1{\mathrm{m}}^{-2}=10^{-20}{\mathring{\mathrm{A}}}^{-2}$.

Use the water data to determine the common oxygen contribution rather than assume identical molar volumes. Ordinary water has molecular number densityso $b_O={\rho }_b^{H_{2}O}/n_H-2b_H\simeq 5.81\times 10^{-15}\mathrm{m}$. Therefore $b_{D_{2}O}=2b_D+b_O\simeq 1.915\times 10^{-14}\mathrm{m}$ andThe assumptions are isotopically pure waters, additive nuclear scattering lengths, the same oxygen isotope contribution, and molar masses approximately $18$ and $20\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Equal molar volumes would give the rougher estimate $20/18=1.11\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$.

Assuming additive liquid volumes, the scattering length density is linear in D${}_2$O volume fraction ${\varphi }_D$:where all scattering length densities are in $10^{-6}{\mathring{\mathrm{A}}}^{-2}$. The $100\mathrm{g}$ of heavy water occupies about $90.99\mathrm{c}{\mathrm{m}}^3$. The required ordinary-water volume isThus add approximately $147\mathrm{g}$ of H${}_2$O. Volume fraction, rather than mass fraction, enters this contrast-matching equation. Isotope exchange to form HDO does not change the total nuclear scattering-length balance in the additive-volume approximation.

Without the film, a sharp silicon-water boundary gives the usual Fresnel-like reflectivity governed by its scattering-length-density contrast. If the contrast-matched mixture from part (iii) is used, that ideal interface signal is strongly suppressed.

Polyethylene has a different scattering length density from both matched media. Its two interfaces generate interference oscillations, or Kiessig fringes, in reflectivity versus normal momentum transfer $q_z$. Their approximate spacing isfor $d=100\mathrm{nm}=1000\mathring{\mathrm{A}}$. Fit several fringes, with refraction, roughness and instrument resolution included if necessary, to determine the thickness accurately. Even with unmatched water the film changes the reflection envelope and adds thickness-dependent interference.