Advertisements
Advertisements
Question
A face centred cube (FCC) consists of how many atoms? Explain
Advertisements
Solution
Face-centred cubic lattice (fcc):
1) In face-centred cubic unit cell, eight constituent particles (spheres) are present at eight corners of unit cell. Six constituent particles (spheres) are present at centres of six faces

2) A constituent particle present at a corner is shared by eight neighbouring unit cells. Its contribution to a unit cell is only 1/8. Thus, the number of atoms present at corners per unit cell
= 8 corner atoms x 1/8 atom per unit cell = 1
3) A constituent particle present at the centre of a face is shared by two neighbouring unit cells. Its contribution to a unit cell is only 1/2.
The number of atoms present at faces per unit cell
= 6 atoms at the faces x 1/2 atom per unit cell = 3
4) The total number of atoms per unit cell = 1 + 3 = 4
Thus, a face-centred cubic unit cell has 4 atoms per unit cell.
APPEARS IN
RELATED QUESTIONS
A unit cell of iron crystal has edge length 288 pm and density 7.86 g.cm-3. Find the number of atoms per unit cell and type of the crystal lattice.
Given : Molar mass of iron = 56 g.mol-1; Avogadro's number NA = 6.022 x 1023.
How many atoms constitute one unit cell of a face-centered cubic crystal?
Gold occurs as face centred cube and has a density of 19.30 kg dm-3. Calculate atomic radius of gold. (Molar mass of Au = 197)
Face centred cubic crystal lattice of copper has density of 8.966 g.cm-3. Calculate the volume of the unit cell. Given molar mass of copper is 63.5 g mol-1 and Avogadro number NA is 6.022 x 1023 mol-1
An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m−3, what is the nature of the cubic unit cell?
What is the coordination number of atoms in a cubic close-packed structure?
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Explain with reason sign conventions of ΔS in the following reaction
N2(g) + 3H2(g) → 2NH3(g)
An element has atomic mass 93 g mol−1 and density 11.5 g cm–3. If the edge length of its unit cell is 300 pm, identify the type of unit cell. (NA = 6.023 × 1023 mol−1)
Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a f.c.c. structure. (Atomic mass of Al = 27 g mol–1)
The density of silver having an atomic mass of 107.8 g mol- 1 is 10.8 g cm-3. If the edge length of cubic unit cell is 4.05 × 10- 8
cm, find the number of silver atoms in the unit cell.
( NA = 6.022 × 1023, 1 Å = 10-8 cm)
Number of types of orthorhombic unit cell is ___________.
The number of atoms per unit cell in a body centered cubic structure is ____________.
An atom located at the body center of a cubic unit cell is shared by ____________.
TiCl has the structure of CsCl. The coordination number of the ions in TiCl is ____________.
An element forms a cubic unit cell with edge length 405 pm. Molar mass of this element is 2.7 × 10−2 kg/mol and its density is given as 2.7 × 103 kg/m3. How many atoms of these elements are present per unit cell?
A metal has a body-centered cubic crystal structure. The density of the metal is 5.96 g/cm3. Find the volume of the unit cell if the atomic mass of metal is 50.
An element (atomic mass 100 g/mol) having bcc structure has unit cell edge 400 pm. The density of element is (No. of atoms in bcc, Z = 2).
The number of atoms contained in a fcc unit cell of a monoatomic substance is ____________.
Which of the following metal(s) show(s) hexagonal close-packed structure (hcp) and which show face-centered cubic (fcc) structure?
The density of a metal which crystallises in bcc lattice with unit cell edge length 300 pm and molar mass 50 g mol−1 will be:
An element with atomic mass 100 has a bcc structure and edge length 400 pm. The density of element is:
The percentage of empty space in a body centred cubic arrangement is ______.
Match the type of unit cell given in Column I with the features given in Column II.
| Column I | Column II |
| (i) Primitive cubic unit cell | (a) Each of the three perpendicular edges compulsorily have the different edge length i.e; a ≠ b ≠ c. |
| (ii) Body centred cubic unit cell | (b) Number of atoms per unit cell is one. |
| (iii) Face centred cubic unit cell | (c) Each of the three perpendicular edges compulsorily have the same edge length i.e; a = b = c. |
| (iv) End centred orthorhombic cell | (d) In addition to the contribution from unit cell the corner atoms the number of atoms present in a unit cell is one. |
| (e) In addition to the contribution from the corner atoms the number of atoms present in a unit cell is three. |
Percentage of free space in body centred cubic unit cell is
If a represents the edge length of the cubic systems, i.e. simple cubic, body centred cubic and face centered cubic, then the ratio of the radii of the sphere in these system will be:-
