Niobium forms bcc structure the density of niobium is 8.55 g/cm3 and length of unit cell edge is 330.6 pm . How many atoms and units cells are present in 0.5 g of niobium ?
Best answer
Answer: In 0.5 g of niobium there are \(1.62\times10^{21}\) unit cells and \(3.24\times10^{21}\) atoms.
Given: bcc so \(n = 2\) atoms per unit cell, \(\rho = 8.55\ \text{g cm}^{-3}\), \(a = 330.6\ \text{pm} = 330.6\times10^{-10}\ \text{cm}\), mass \(= 0.5\ \text{g}\)
Step 1: Volume of 0.5 g of Nb
\(V = \dfrac{m}{\rho} = \dfrac{0.5}{8.55} = 5.848\times10^{-2}\ \text{cm}^3\)
Step 2: Volume of one unit cell
\(a^3 = (330.6\times10^{-10})^3 = 3.613\times10^{-23}\ \text{cm}^3\)
Step 3: Number of unit cells
\(\dfrac{V}{a^3} = \dfrac{5.848\times10^{-2}}{3.613\times10^{-23}} = 1.62\times10^{21}\)
Step 4: Number of atoms
Each bcc unit cell has 2 atoms, so atoms \(= 2\times1.62\times10^{21} = 3.24\times10^{21}\)
Answer: 0.5 g of niobium contains about \(1.62\times10^{21}\) unit cells and \(3.24\times10^{21}\) atoms.
Step 1: volume of one unit cell
\(a = 330.6\ \text{pm} = 330.6\times10^{-10}\ \text{cm}\)
\[a^3 = (3.306\times10^{-8})^3 = 3.613\times10^{-23}\ \text{cm}^3\]
Step 2: mass of one unit cell
\[m = \rho a^3 = 8.55 \times 3.613\times10^{-23}\]
\[m = 3.089\times10^{-22}\ \text{g}\]
Step 3: number of unit cells in 0.5 g
\[\frac{0.5}{3.089\times10^{-22}} = 1.62\times10^{21}\]
Step 4: number of atoms
A bcc unit cell has \(8\times\tfrac18 + 1 = 2\) atoms, so
\[2 \times 1.62\times10^{21} = 3.24\times10^{21}\ \text{atoms}\]
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