Heavy-Fermion Properties of Yb2Pd2SnH≈2

Silvie Maskova-Cerna, Ernst Bauer, Mauro Giovannini, Ladislav Havela
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Abstract

A hydride of Yb2Pd2Sn could be synthesized with approximately 2 H atoms per f.u. The hydrogenation leads to a volume expansion while preserving the tetragonal symmetry (P4/mbm). The lattice reaction is strongly anisotropic, and the 5% expansion in c is partly compensated by the 0.5% compression in a. The hydride is paramagnetic at least down to 0.5 K. Yb remains at or very close to the 3+ (4f13) state, as in Yb2Pd2Sn. Specific heat C/T vs. T shows an upturn existing already in Yb2Pd2Sn, but it is much more pronounced in the hydride (1.8 J/mol f.u. K2 for T → 0, i.e., more than twice higher than in its precursor). This is interpreted as lowering the Kondo temperature due to H bonding.
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Yb2Pd2SnH≈2的重费米子性质
Yb2Pd2Sn的氢化物可以以每f.u约2个H原子的速度合成,氢化导致体积膨胀,同时保持了四方对称(P4/mbm)。晶格反应具有很强的各向异性,c中5%的膨胀被a中0.5%的压缩部分补偿。氢化物至少在0.5 K以下是顺磁性的。Yb保持在或非常接近3+ (4f13)状态,如Yb2Pd2Sn。比热C/T比T在Yb2Pd2Sn中已经出现了上升,但在氢化物中更为明显(T→0时为1.8 J/mol f.u K2,即比其前体高两倍以上)。这被解释为由于氢键作用降低了近藤温度。
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