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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
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(0.389 nm)2 = 0.151 nm2
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