Lattice dynamics of Eu from nuclear inelastic scattering and first-principles calculations

We report on combined determination of the density of phonon states of an epitaxial Eu metallic single crystal by in situ ${}^{151}\text{E}\text{u}$ nuclear inelastic scattering and by first-principles phonon calculations. Thermodynamic (specific heat and vibrational entropy) and elastic (mean atomic displacement, mean force constant, and sound velocity) properties of Eu were obtained. The very good agreement between the experiment and theory indicates that in general, the lattice dynamics of the $4f$ metals can be reliably investigated by the ab initio approach applying the direct method.

Figure 1

(Color online) (a) X-ray diffraction scan $\left(\lambda =0.86\text{Å}\right)$ with the assigned Eu reflections. Peaks arising from the substrate are also visible. (b) Time spectrum (open circles) of the coherent nuclear forward scattering recorded at 100 K and at an incidence angle of $0.1°$, which corresponds to the critical angle. The solid (red) line is a simulation according to the theory.

Figure 2

(Color online) Energy dependence of nuclear inelastic x-ray absorption in the 200-nm-thick bcc (110) Eu metallic film measured at 100 K. The inset shows a fit of the instrumental function (solid circles) with a Gaussian distribution (solid/red line) with a full width at half maximum of 1.45(2) meV.

Figure 3

(Color online) Phonon DOS for bcc europium. The closed circles and the thin solid (black) line mark the DOS derived from the energy dependence of the nuclear inelastic absorption measured at 100 K. The solid (red) line represents the calculated DOS from Fig. 4b convoluted with a Gaussian function accounting for the instrumental broadening.

Figure 4

(Color online) Results of the calculations with PHONON code (Ref. 36) for bulk bcc europium: (a) phonon-dispersion relations within the first Brillouin zone and (b) phonon DOS. The high-symmetry points in units of $2\pi /a$ are $\Gamma =\left(0,0,0\right)$, $N=\left(\frac{1}{2},\frac{1}{2},0\right)$, $P=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$, and $H=\left(0,0,1\right)$.