Advertisements
Advertisements
प्रश्न
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Advertisements
उत्तर
Suppose, the edge length of the unit cell = a
Number of atoms present per unit cell = Z
Atomic mass of the element = M
For a cubic unit cell,
Volume of the unit cell = a3 ...(i)
Density of the unit cell
ρ' = `"Mass of unit cell"/"Volume of unit cell"` ...(ii)
Mass of unit cell = Number of atoms per unit cell × Mass of one atom
= Z × m ...(iii)
where m is the mass of a single atom. This is given by
`m = "Atomic mass"/"Avogadro’s number"`
`m = M/N_A` ...(iv)
Substituting the value of m in eq. (iii), we have
Mass of unit cell = `Z xx M/N_A`
Substituting the corresponding values in eq. (ii), the density of the unit cell (ρ') is given by
ρ' = `"Mass of unit cell"/"Volume of unit cell"`
ρ' = `(Z xx M/N_A)/a^3`
or, ρ' = `(Z xx M)/(a^3 xx N_A)`
The density of a crystal (ρ) is the same as the density of its unit cell (ρ'), i.e., ρ' = ρ. Therefore the density ρ of a crystal is given by
`rho = (Z xx M)/(a^3 xx N_A)`
∴ `M = (rho xx a^3 xx N_A)/(Z)`
संबंधित प्रश्न
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 27 g mol−1 forms a cubic unit cell with edge length 4.05 ✕ 10−8 cm. If its density is 2.7 g cm−3, what is the nature of the cubic unit cell?
Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.
A face centred cube (FCC) consists of how many atoms? Explain
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)
An element 'X' (At. mass = 40 g mol-1) having f.c.c. the structure has unit cell edge length of 400 pm. Calculate the density of 'X' and the number of unit cells in 4 g of 'X'. (NA = 6.022 × 1023 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)
What is the total number of atoms per unit cell in a face-centered cubic structure?
An atom located at the body center of a cubic unit cell is shared by ____________.
A substance forms face-centered cubic crystals. Its density is 1.984 g/cm3 and the length of the edge of the unit cell is 630 pm. Calculate the molar mass in g/mol?
Sodium metal crystallises in a body-centred cubic lattice with a unit cell edge of 4.29 Å. The radius of the sodium atom is approximately ______.
The number of atoms contained in a fcc unit cell of a monoatomic substance is ____________.
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. |
The correct set of quantum numbers for 3d subshell is
The ratio of number of atoms present in a simple cubic, body-centred cubic and face-centred cubic structure are, respectively ______.
