Question

An element A (Atomic weight $ = 100{\text{ amu}} $ ) having BCC structure has unit cell edge length of $ 400{\text{ pm}} $ . Calculate the density (in g/cc) of A?
(A) $ 5.188 $
(B) $ 7.188 $
(C) $ 4.144 $
(D) $ 10.144 $

Answer 1

Hint: To answer this question, you must recall the formula for calculating the density of a unit cell. You must also recall the method for finding the number of atoms in a unit cell. The number of atoms in a BCC unit cell is 2.
Formula used: $ d = \dfrac{{n \times M}}{{{N_A} \times {a^3}}} $
Where, $ d $ denotes the density of the unit cell
 $ n $ represents the number of atoms present in a unit cell
 $ M $ represents the molecular mass of the given substance.
And, $ a $ represents the length of edge of the cubic unit cell.

Complete step by step solution
The given element A is said to have a BCC structure, that is, a body centred cubic structure. We know that the particles (atoms or molecules) in a body centred cubic unit cell lie at the corners of the unit cell and one particle at the body center of the cell. Thus, the number of particles present in the BCC unit cell is 2.
We get the value of $ n = 2 $
We know that the atomic weight measured in atomic mass units is equal to the molar mass of the element in grams. So molar mass $ = M = 100{\text{ g}} $
We know the formula for density of unit cell as, $ d = \dfrac{{n \times M}}{{{N_A} \times {a^3}}} $
Substituting the values, we get,
 $ d = \dfrac{{2 \times 100}}{{6.023 \times {{10}^{23}} \times {{\left( {400 \times {{10}^{ - 10}}} \right)}^3}}} $
 $ \Rightarrow d = 5.188{\text{ g c}}{{\text{m}}^{ - 3}} $
Thus, the correct answer is A.

Note
It should be known that in a body centred cubic unit cell, atoms are present at the corners and the centre of the cube. So, we can write that,
 $ {n_c} = $ number of atoms in the corners of the BCC unit cell $ = 8 $
 $ {n_f} = $ number of atoms on the six faces of the BCC unit cell $ = 0 $
 $ {n_i} = $ number of atoms completely inside the BCC unit cell $ = 1 $
 $ {n_e} = $ number of atoms at the edges of the BCC unit cell $ = 0 $
Thus, the total number of atoms in a body centred unit cell is
 $ n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4} $
Substituting the values, we get
 $ n = \dfrac{8}{8} + \dfrac{0}{2} + \dfrac{1}{1} + \dfrac{0}{4}{\text{ }} $
 $ n = 1 + 0 + 1 + 0{\text{ }} $
Thus, $ n = 2 $.