If edge of a bcc crystal of an element is a cm,M is the atomic mass and `N_(0)` is Avogadro number then density of the crystal is

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

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 ?

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.)

Each edge of a cubic unit cell is 400pm long. If atomic mass of the elements is 120 and its desity is 6.25g//cm^(2) , the crystal lattice is: (use N_(A)=6 xx 10^(23))

In a cubic lattice each edge of the unit cell is 400 pm. Atomic weigth of the element is 60 and its density is 625g//c.c. Avogadro number = 6times10^(23) .The crystal lattice is

Derive an expression for the density of a cubic crystal whose edge is 'a' pm and contains z atoms per unit cell. The atomic mass of substance may be taken as M.

Na and Mg crystallize in bcc- and fcc-type crystals, respectively, then the number of atoms of Na and Mg present in the unit cell of their respective crystal is

A crystalline solid of pure substance has a face - centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is "8 g cm"^(-3) , then the number of atoms present in 128 g of the crystal is

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8 g cm^(-3) , then the number of atoms present in 256g of the crystal is N xx 10^(24) . The value of N is

Element X crystallizes in 12 co - ordination fcc lattice. On applyng high temperature it changes to bcc lattice. Find the ratio of the density of the crystal lattice before and and after applying high temperature