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´ó¼ÒºÃ£¡ÒÔÏÂÊÇÎÒ¼ÆËãPd¶þÊ®ÃæÌå·ÖÉ¢¶ÈµÄ·½·¨¡£Çë´ó¼Ò¼ì²éÏÂÊÇ·ñ¶Ô²»¡£»òÕßÓÐÆäËü¼ÆËã·½·¨Çë´ó¼Ò¶à¶àÖ¸½Ì¡£Ð»Ð»À²¡£ Calculation dispersion of Pd icosahedron The Pd icosahedron sample has an average edge length of 14.5 nm. Its volume is: (15+5¡Áadical sign(5)) /12 ¡Á(14.5 nm)3 = 6651.17 nm3 Pd has the face-centered-cubic structure with a lattice constant of 0.389 nm. The volume of a unit cell is : (0.389 nm)3 = 0.0589 nm3, Each unit cell contains 4 Pd atoms. The number of Pd atoms in a single Pd icosahedron is: (6651.17 nm3) / (0.0589 nm3) ¡Á4 = 451692, Its total surface area is: (14.5 nm)2 ¡Á( 5 adical sign(3))= 1820.8 nm2, Each two-dimensional unit cell on the (111) facet contains two Pd atoms. Its area is (0.389 nm)2 = 0.151 nm2 The number of the surface Pd atoms on a single Pd icosahedron is: 1820.8 nm2/(0.151 nm2) ¡Á 2 = 24117 The dispersion of a single Pd icosahedron is: 24117/451692 = 0.05 Ö÷ÒªÓÐÒÉÎʵĵط½¾ÍÊÇÉÏÃæ±êÏ»®Ïߵĵط½¡£²»ÖªµÀÕâÑùËã¶Ô²»?лл´ó¼Ò¿©¡£ |
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