Question

The density of nickel (Face centred cubic cell) is 8.94g/ $ c{m^3} $  at 20 $ ^\circ C $  . What is the radius of the atom? (Atomic weight: Ni = 59)

(A) 0.124 nm

(B) 0.136 nm

(C) 0.149 nm

(D) 0.110 nm

Answer 1

Hint: As the name suggests, a Face Centred Cubic (FCC) lattice structure has atoms on all the sides of the cubic structure as well as all the vertices. There is no atom present in the centre of the cube. This brings the total count of the total number of atoms present in a Face Centred Cubic structure to 4.

Complete Step-by-Step Solution:

Now, in order to calculate the diameter of the atom, we need to follow a few steps:

Calculate the edge length of the FCC cube:

We know that,

Density = (mass)/(volume)

Applying this concept for calculating the density of the unite cell, we get,

Density =  $ \dfrac{{(M \times z)}}{{(N \times {a^3})}} $ 

=  $ \dfrac{{(molecular{\text{ }}weight){\text{ (}}number{\text{ }}of{\text{ }}atoms{\text{ }}per{\text{ }}cell){\text{ }}}}{{{{(edge{\text{ }}length)}^3}{\text{ (}}Avogadro{\text{ }}number)}} $ 

=  $ \dfrac{{59 \times 4}}{{{a^3} \times 6.022 \times {{10}^{23}}}} $ 

Hence,

 $ {a^3} $  =  $ \dfrac{{59 \times 4}}{{density \times 6.022 \times {{10}^{23}}}} $ 

 $ {a^3} $  =  $ \dfrac{{59 \times 4}}{{8.94 \times 6.022 \times {{10}^{23}}}} $ = 4.83  $  \times {10^{ - 23}} $ 

Hence,

a =  $ \sqrt[3]{{4.83}} \times {10^{23}} $ = 3.64 $  \times {10^{ - 8}} $ cm

Calculate the radius of the atom:

We know that, the relation between the radius of the atom and the edge length of the unit cell can be given by –

 $ \sqrt {2a} = 4r $ 

Hence, r =  $ \dfrac{{\sqrt 2 a}}{4} $ 

r =  $ \dfrac{{\sqrt 2 \times 3.64 \times {{10}^{ - 8}}}}{4} $ = 1.28 $  \times {10^{ - 8}} $ cm = 0.128 nm  $  \approx  $ 0.124 nm

Hence the diameter of the atom can be calculated as:

d = 2r = 2(1.28 $  \times {10^{ - 8}} $ ) = 2.48 $  \times {10^{ - 8}} $ cm

Hence, Option (A) is the correct option.

Note: The formulae to calculate the radii of the constituent atom is different for different lattice structures but can be derived in a similar manner. Although it is advisable to memorise these formulae for the ease of calculation.