Derive an expression for the calculation of density of the cubic crystal of an element whose edge is 'a' pm and atomic mass is M.

Derive an expression for density of a cubic crystal from the edge of the cubic crystal of an element in terms of SI units.

Density of a unit cell is same as the density of the substance. If the density of the substance is known, number of atoms or dimensions of the unit cell can be calculated . The density of the unit cell is related to its mass(M), no. of atoms per unit cell (Z), edge length (a in cm) and Avogadro number N_A as : rho = (Z xx M)/(a^3 xx N_A) An element crystallizes in a structure having a fcc unit-cell an edge 100 pm. If 24 g of the element contains 24 xx 10^(23) atoms, the density is

Determine the type of cubic lattice to which a crystal belongs if the unit cell edge length is 290 pm and the density of crystal is 7.80g cm^(-3) . (Molar mass = 56 a.m.u.)

A element crystallises in a bcc structure. The edge length of its unit cell is 288 pm. If the density of the crystal is 7.3 g cm^(-3) , what is the atomic mass of the element ?

Knowing the edge length of a cubic crystal of an element and its. Density , how will your find the value of Avogadro's number ?

If the unit cell edge length of NaCl crystal is 600 pm, then the density will be

An element has a body centered cubic (bcc) structure with a cell edge of 288pm. The atomic radius is: