If an element crystallizes as a simple cube, what is the volume of an element provided its density is 1.5 g/cm3 and atomic mass of the element is 63?
Simple cube has 1 atom
Therefore volume of cube i.e. a3 = 6.97 x 10-23 cm3
Mass of an atom in an unit cell is calculated by which of the following formula if M is the molar mass:
(Molar mass = M) (1 molar of atom = M 6.023×1023)( atom = M/Na)
In the formula to calculate the density of a unit cell what is z ?
Volume of a body centred cubic unit cell is
In a body centered cubic unit cell ,√3a = 4r a= 4r/√3 where a = side of the cube volume = a3 = (4r/√3)3.
Density of a unit cell is equal to mass of unit cell divided by
Volume of unit cell, V = a3
=> Density of unit cell = mass of unit cell /volume of unit cell
Length a in unit cell is along
The number of atoms (z) in simple cubic unit cell is
The face-centered cubic (fcc) has a coordination number of 12 and contains 4 atoms per unit cell. The body-centered cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell. The simple cubic has a coordination number of 6 and contains 1 atom per unit cell.
X-ray diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of 3.608 x 10-8 cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm3, calculate the atomic mass of copper.
In case of fcc lattice,number of atoms per unit cell,x=4 atoms
∴ Atomic mass of copper =63.1u
An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will be
Length of the edge , a = 290 pm =290 x 10-10 cm
Volume of unit cell = ( 290 x 10-10 cm )3 = 24.39 x 10-24 cm3
Since it is bcc arrangement,
Number of atoms in the unit cell, Z = 2
Atomic mass of the element = 50
Mass of the atom = atomic mass/ Avogadro number = M/No = 50/6.02 x 1023
Mass of the unit cell = Z x M/No = 2 x 50/6.02 x 1023 = 100/6.23 x 1023
Therefore , density = mass of unit cell / volume of unit cell
= 100/6.023 x 1023 x 24.39 x 10-24 = 6.81 g cm-3
The number of octahedral voids (s) per atom present in a cubic close-packed structure is
No of atoms per unit cell is 4,,it means it contain 4 octahedral voids and 8 tetrahedral voids
Total no of octahedral voids per atom is 1