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Question: The density of copper is 8.95\(g\text{ c}{{\text{m}}^{-3}}\). It has a face centred cubic structure....

The density of copper is 8.95g cm−3g\text{ c}{{\text{m}}^{-3}}. It has a face centred cubic structure. What is the radius of the copper atom? (Atomic mass of copper = 63.5 g/mol, NA=6.02×1023mol−1{{N}_{A}}=6.02\times {{10}^{23}}mo{{l}^{-1}})

Explanation

Solution

To solve this, firstly find out the mass of a single copper atom and then multiply it by the total atoms in the fcc unit cell to get the total mass. Use this to find out the volume. You can calculate the edge length by taking the cubic root of volume. And then finally use the relation 2a=4r\sqrt{2}a=4r to find the radius.

Complete step by step solution:
We know that Fcc lattice is the abbreviation used for face centred cubic lattice. In the face centred cubic lattice, atoms are arranged in each corner and at the centre of each face.
Firstly, let us calculate the mass of one copper atom by dividing its atomic mass by Avogadro’s number.
We can write that-
Mass of one copper atom = Atomic mass of copperNA=63.5 g/mol6.02×1023mol−1=1.054×10−22g\dfrac{Atomic\text{ mass of copper}}{{{N}_{A}}}=\dfrac{63.5\text{ }g/mol}{6.02\times {{10}^{23}}mo{{l}^{-1}}}=1.054\times {{10}^{-22}}g
Now, in the fcc lattice there are 4 atoms in a unit cell. Therefore, we can write that the total mass of the unit cell is 4×1.054×10−22g4\times 1.054\times {{10}^{-22}}g
Now we will calculate the volume of the unit cell using the formula- density=mass ÷\div volume
The density is given to us in the question and we have calculated the mass of the unit cell. So, putting the values in the above formula we will get-
volume=MassDensity=4×1.054×10−22g8.95g cm−3=4.71×10−23cm3volume=\dfrac{Mass}{Density}=\dfrac{4\times 1.054\times {{10}^{-22}}g}{8.95g\text{ c}{{\text{m}}^{-3}}}=4.71\times {{10}^{-23}}c{{m}^{3}}
Now, we know that for fcc unit cell, the relation between edge length and radius is- 2a=4r\sqrt{2}a=4r
Or, we can write that a = 22r2\sqrt{2}r
Also, we know that volume = a3{{a}^{3}} = 4.71×10−23cm34.71\times {{10}^{-23}}c{{m}^{3}}. Therefore, a = 3.61×10−8cm3.61\times {{10}^{-8}}cm
So, putting the values in the edge length-radius relation we can write that-
3.61×10−8cm=22r or,r=3.61×10−8cm22=1.2768×10−8cm \begin{aligned} & 3.61\times {{10}^{-8}}cm=2\sqrt{2}r \\\ & or,r=\dfrac{3.61\times {{10}^{-8}}cm}{2\sqrt{2}}=1.2768\times {{10}^{-8}}cm \\\ \end{aligned}

Therefore, the required radius of copper atom is 1.2768×10−8cm1.2768\times {{10}^{-8}}cm

Note: We know that a lattice is a network which gives us the three dimensional arrangement of atoms. There are 14 different Bravais lattices that we study in case of solids. The body centred and the cubic close packed structures are also Bravais lattice. BCC is one of the primitive lattices and CCP is just like the face centred Bravais lattice. Face Centered Cubic i.e. fcc and cubic close packed i.e. ccp, these are two different names for the same lattice. We can think of this cell as being made by inserting another atom into each face of the simple cubic lattice - hence the "face-centred cubic" name.