Gold nano particles are being studied by material scientists for a variety of us

Question

Gold nano particles are being studied by material scientists for a variety of uses, including the treatment of cancer and other diseases. Consider specifically a spherical gold nano particle with a diameter of 33.0 nm. The atomic weight of gold is 196.7 g/mol and the atomic radius is .144nm. The crystal structure is FCC. a. Calculate the lattice parameter of the unit cell. . b. Use calculated lattice parameter to find the density of gold. c. Assuming this particle is a single crystal, how many unit cells are lined up across the diameter of the particle? d. Calculate the mass of the particle. e. Approximately how many atoms does this particle contain?

Explanation / Answer

for lattice parameter to me a and atomic radius of gold to be r we can say a^2 + a^2 = (4r)^2 ... for FCC as length of the diagonal will be 4 times atomic radius hence solving above we get a= 2(2r)^(1/2) putting value of r we get a=0.4073 ...(a) This is the lattice parameter required now there is an atom at each corner that is in a total of 8 cells. so 1/8 of each corner atom is in a particular cell. there are 8 corners so 8 x 1/8 = 1 atom per cell from the corners. There are 6 sides. each side has an atom that exists in two cells.. so atom is 1/2 in each cell. 6 x 1/2 = 3 atoms for our unit cell... 1 + 3 = 4 atoms / cell volume occupied by the nano particle will be a*a*a =6.7568*10^(-29) m^3 mass of that nano particle = 4 * (mass of one atom of gold) =4 * (atomic mass/ Avagadro number) =4 * 196.7/(6.022 * 10^23) =1.3065*10^(-21) gram density = mass/volume = 1.3065*10^(-21) / 6.7568*10^(-29) =19336076.25 gm/m^3 =19336.076 kg/m^3 ... (b)

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