Calculate the number of atoms present in 2 gram of crystal which has f

1.	Calculate the number of atoms present in 2 gram of crystal which has face-centred cubic (FCC) crystal lattice having edge length of 100 pm and density 10 g cm-3.
Answer» Given: Density (d) = 10 g cm^-3 Edge length (a) = 100 pm = 100×10^–12m = 100×10¹⁰cm Mass of crystal = 2 g To find: Number of atoms Formula: Density =\(\frac{Mass}{Volume}\) Density=\(\frac{Mass}{Volume}\) ∴ Volume =\(\frac{Mass}{Density}\) ∴ Volume = \(\frac{2\,g}{10\,g\,cm^{-3}}=0.2\,cm^3\) Volume of unit cell = a³ = (100×10^–10cm)3 = 1×10^-24 cm³ Number of unit cells in 2 g of crystal =\(\frac{total\,volume}{volume\,of\,unit\,cell}\) =\(\frac{0.2\,cm^3}{1\times10^{-24}cm^3}\) = 0.2×10²⁴unit cells ∵ The given unit cell is of fcc type, therefore, it contains 4 atoms. ∴ 0.2×10²⁴ unit cells will contain 4×0.2×10²⁴ = 0.8×10²⁴ atoms = 0.8×10²⁴ atoms = 8×10²³ atoms Ans: Number of atoms present in 2 g of given crystal is 8×10²³ atoms.

Calculate the number of atoms present in 2 gram of crystal which has face-centred cubic (FCC) crystal lattice having edge length of 100 pm and density 10 g cm-3.

Answer»

Given: Density (d) = 10 g cm^-3

Edge length (a) = 100 pm = 100×10^–12m = 100×10¹⁰cm

Mass of crystal = 2 g

To find: Number of atoms

Formula: Density =\(\frac{Mass}{Volume}\)

Density=\(\frac{Mass}{Volume}\)

∴ Volume =\(\frac{Mass}{Density}\)

∴ Volume = \(\frac{2\,g}{10\,g\,cm^{-3}}=0.2\,cm^3\)

Volume of unit cell = a³ = (100×10^–10cm)3 = 1×10^-24 cm³

Number of unit cells in 2 g of crystal =\(\frac{total\,volume}{volume\,of\,unit\,cell}\)

=\(\frac{0.2\,cm^3}{1\times10^{-24}cm^3}\)

= 0.2×10²⁴unit cells

∵ The given unit cell is of fcc type, therefore, it contains 4 atoms.

∴ 0.2×10²⁴ unit cells will contain 4×0.2×10²⁴ = 0.8×10²⁴ atoms

= 0.8×10²⁴ atoms = 8×10²³ atoms

Ans: Number of atoms present in 2 g of given crystal is 8×10²³ atoms.