Question

Nickel has a face-centered cubic unit cell. The density of nickel is 6.84 g/cm³. Calculate a value for the atomic radius of nickel.

Short answer

The atomic radius of Nickel is$1.89\times {10}^{-8}\mathrm{cm}.$

Step-by-step solution

Calculating the mass of a single Ni atom

Density = mass of unit cell/ volume of a unit cell. Since Nickel is FCC, the Unit cell contains 4 Ni atoms per cell.

Mass of unit cell = Mass of 4 Ni atoms

The mass of Nickel is 58.6 g/mol.

That means 1 mole of nickel contains $6.023\times {10}^{23}\mathrm{Ni}\mathrm{atoms}.$

So, the average mass of a single Ni atom can be calculated by dividing 58.6 gm with Avogadro's Number.

Therefore,

Mass of the single Ni atom =$\frac{58.6\mathrm{gm}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}}$

As the unit cell is FCC, it has four atoms in its unit cell.

Mass of the unit cell = 4 x Mass of a single Ni atom

Calculating atomic radius of nickel

It is given that the density of Nickel is 6.84 g/cc.

$\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{unit}\mathrm{cell}={\left(\mathrm{edge}\mathrm{Length}\right)}^3$

$\mathrm{Density}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{unit}\mathrm{cell}}{{\mathrm{a}}^3}$

In FCC unit cell, the relation between edge length and atomic radius is:$\mathrm{a}=2\mathrm{r}$

So,

role="math" localid="1664131578264" $$\begin{gathered} 6.84\mathrm{gm}/{\mathrm{cm}}^3=\frac{4\times 58.6\mathrm{gm}/{\mathrm{cm}}^3}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}\times {\mathrm{a}}^3} \\ {\mathrm{a}}^3=\frac{4\times 58.6\mathrm{gm}/{\mathrm{cm}}^3}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}\times 6.84\mathrm{gm}/{\mathrm{cm}}^3} \\ \mathrm{a}=3.79\times {10}^{-8}\mathrm{cm} \end{gathered}$$

Therefore, the edge length is role="math" localid="1664131707069" $3.79\times {10}^{-8}\mathrm{cm}.$

The atomic radius of the Nickel = a/2 =$1.89\times {10}^{-8}\mathrm{cm}.$