 Solution

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.