A metal exist as face centered cubic crystals. If the atomic radius is `100sqrt(2)`pm and the density of metal is `12,500 kg//m^(3)`, the metal is :9Given : Ca = 40, Co= 58.9 , Sn = 119.8, Pb= 208.4, N_(A) = 6 xx 10^(23)`

Aluminium metals forms a cubic face centred close packed crystal structure. Its atomic radius is 125xx10^(-12) m. The length of the side of the unit cell is

Copper crystallises with face centred cubic unit cell . If the radius of copper atom is 127.8 pm , calculate the density of copper metal. (Atomic mass of Cu = 63.55 u and Avogadro's number N_(A) = 6.02 xx 10^(23) mol^(-1) )

Copper crystallises with face centred cubic unit cell . If the radius of copper atom is 127.8 pm , calculate the density of copper metal. (Atomic mass of Cu = 63.55 u and Avogadro's number N_(A) = 6.02 xx 10^(23) mol^(-1) )

Copper crystallises with face-centred cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal. (Atomic mass of Cu = 63.55 g/mol and Avogadro's number N_(A) = 6.02 xx 10^(23) mol^(-1) )

A metallic element exists in cubic lattice. Each edge of unit cell is 4 Å. The density of metal is 6.25 g//m^(3) . How many unit cells will be present in 100 g of metal?

A metallic element exists in cubic lattice. Each edge of unit cell is 4 Å. The density of metal is 6.25 g//m^(3) . How many unit cells will be present in 100 g of metal?

Copper crystallises with face centred cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal. (Atomic mass of Cu= 63.55 u and Avogadro's number, N_(A)=6.02xx10^(23)"mol"^(-1) )