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The main characteristic peaks identify nickel and phosphorus: the spectrum has a strong Ni $K\alpha$ peak near $7.48\mathrm{keV}$, the Ni $L$ region near $0.85\mathrm{keV}$ and P $K\alpha$ near $2.01\mathrm{keV}$. Weaker associated lines, including Ni $K\beta$ near $8.26\mathrm{keV}$, can accompany them. The continuous background is bremsstrahlung from decelerated electrons.

Small carbon and oxygen signals at low energy can arise from contamination, surface oxide or preparation materials; a copper contribution can arise from the adjacent substrate within the finite interaction volume. The Cu $L$ region lies close to Ni $L$, so overlap is particularly troublesome at low energy. A weak Cu $K\alpha$ contribution near $8.04\mathrm{keV}$ is better separated, though still close to Ni $K\beta$. At $30\mathrm{kV}$ the beam can excite these K lines.

Peak height is not an atomic fraction: ionization cross-sections, fluorescence yields, absorption, detector response, background and line overlap require correction. The large nickel peak therefore supports the nickel-rich assignment but does not directly give the nickel-to-phosphorus atomic ratio.

Use the supplied polynomial as an effective-atomic-number approximation for the alloys. Interpreting each quoted phosphorus number as an atomic fraction $x_P=N_P/(N_P+N_{\mathrm{Ni}})$,The gold reference yield is $\eta (79)=0.486996$. Substitution gives:

The ordering is gold brightest, then copper, NiP${}_z$, NiP${}_x$ and NiP${}_y$. The phosphorus-rich layer has the lowest effective atomic number.

There is a wording ambiguity in “atomic fraction of P relative to Ni.” If those numbers instead mean $r=N_P/N_{\mathrm{Ni}}$, first convert $x_P=r/(1+r)$, giving $\bar{Z}=(28+15r)/(1+r)$. The three alloy contrasts then become $0.4243$, $0.4361$, $0.4129$, respectively. The qualitative ordering is unchanged. Both calculations explicitly use $\eta (\bar{Z})$; this simple alloy approximation is not a full electron-transport simulation.

Assume a homogeneous Fe–Cr–Ni alloy, normal incidence, independent scattering events and the stated effective scattering cross-sections. Neglect absorption, other constituents and changes in neutron energy within the thin sheet. The atomic fractions are not supplied; take $x_{\mathrm{Fe}}=0.70$, $x_{\mathrm{Cr}}=0.20$, $x_{\mathrm{Ni}}=0.10$ as an explicit representative assumption, not a composition measured or specified in the paper.

The conventional cell of a face-centered cubic lattice containsatoms. The number density is thereforeFor a mixture, the scattering probability per unit path length is ${\Sigma }_s={\sum }_j{n}_j{\sigma }_j=n{\bar{\sigma }}_s$, whereThus ${\Sigma }_s=27.751{\mathrm{m}}^{-1}$. For thickness $t=10^{-3}\mathrm{m}$, the unscattered fraction $U$ satisfies $dU/dz=-{\Sigma }_{s}U$ with $U(0)=1$. The fraction undergoing at least one neutron scattering event isThe thin-sheet approximation gives $P_s\simeq {\Sigma }_{s}t=2.78\%$, consistent to the accuracy of the assumed composition. A simple iron-dominated estimate ${\bar{\sigma }}_s\simeq 3.2$ barns gives essentially the same about $3\%$ scattered conclusion.

For any other assumed composition, use the displayed weighted formula; mass fractions must first be converted to atomic fractions. Since the supplied elemental cross-sections span $2.8$–$4.1$ barns, any mixture of these three elements at the stated number density gives approximately $2.39$–$3.48\%$. This is a bound within the idealized mixture model, not a claim that every such composition is stable austenite.

Use the same atomic fractions and collision assumptions as in part(c). Let $A$ be the irradiated area. The number of atoms in the sheet is $N_t=ntA$, while the number of incident neutrons is $\Phi A$, where $\Phi$ is the neutron fluence.

Counting one primary knock-on atom for each neutron that scatters and using $\nu =430$ displacements per recoil gives the first-collision estimateHence the nominal damage is about $55$ displacements per atom under the assumed composition. In the thin-sheet approximation the density and thickness cancel:The $1.4\%$ difference comes from using first-collision attenuation rather than the linear approximation. Treating repeated collisions after deflection would require angular and energy information not provided; this correction is already smaller than the precision warranted by the composition assumption.

There is also a counting convention in “other atoms”: $430$ counts secondary displacements, while including the displaced primary knock-on atom itself gives $\nu =431$. The corresponding first-collision result is $54.58$ rather than $54.45$ dpa, an increase of only $0.23\%$. Both conventions round to about $55$ dpa. The numerical estimate is explicitly conditional on assigning the stated displacement yield to each scattering event.

A value greater than one is physically possible. Displacements per atom counts displacement events relative to the initial atom inventory: atoms can be displaced repeatedly, and many vacancies and self-interstitials recombine. It does not mean $55$ surviving vacant sites per original lattice site, or that every atom has moved exactly $55$ times. This is a large nominal exposure for which evolving dislocation loops, cavities and irradiation hardening matter; the surviving microstructure additionally depends on irradiation temperature, displacement dose rate, recombination and defect sinks.

An approximate linear interpolation of a substitutional solid solution's lattice parameter with atomic composition. Converting the inferred atomic fraction to mass fraction requires molar masses. Ordering, nonideal mixing, strain and multiple phases can invalidate a simple two-endpoint interpolation.