Question

Calculate the proportion of atoms on the surface atomic layer of Ag spheres with the diameters below given that the thickness of the surface atomic shell is 0.289 nm, stating any assumptions made.

(a) 1 μm (b) 10 nm (c) 5 nm

please explane in detail ,

Answer 1

Silver is FCC elements. So its packing fraction is 74 %

radius of silver atom = 0.1445 nm

Total volume of 1 Ag atom = 4 pi * r³ / 3 = 12.6384*10^-30 m³

a) number of atoms in 1 μm = total volume * packing fraction / volume of 1 atom

= 4.18879 * 10^-18 * 0.74 / 8*12.6384*10^-30 = 3.065*10¹⁰

Total number of atoms on surface = surface area / area of 1 atom

= 4*pi*(0.5*10^-6)² / (pi*1.445²*10^-20)

= 47892147

So, proportion of atoms on the surface atomic layer = 47892147/ 3.065*10^{10 =}0.00156

b)

number of atoms in 10 nm = total volume * packing fraction / volume of 1 atom

= 4.18879 * 10^-24 * 0.74 / 8*12.6384*10^-30 = 3.065*10⁴

Total number of atoms on surface = surface area / area of 1 atom

= 4*pi*(0.5*10^-8)² / (pi*1.445²*10^-20)

= 4789

So, proportion of atoms on the surface atomic layer = 4789/ 3.065*10^{4 =}0.156

c)

number of atoms in 5 nm = total volume * packing fraction / volume of 1 atom

= 4.18879 * 0.5 ³ *10^-24 * 0.74 / 8*12.6384*10^-30 = 3.832*10³

Total number of atoms on surface = surface area / area of 1 atom

= 4*pi*(0.25*10^-8)² / (pi*1.445²*10^-20)

= 1197

So, proportion of atoms on the surface atomic layer = 1197/ 3.832*10^{3 =}0.312

Assumptions: 1. packing fraction = 74%

2. perfect sphere with no defects