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Question: Molybdenum forms body-centred cubic crystals whose density is 10.3\(g.c{m^{ - 3}}\). Calculate the e...

Molybdenum forms body-centred cubic crystals whose density is 10.3g.cm3g.c{m^{ - 3}}. Calculate the edge length of the unit cell. The molar mass of Mo is 95.94g.mol1g.mo{l^{ - 1}}.

Explanation

Solution

First find out the total number of atoms present in a bcc lattice. Then use the formula: d=z×MNA×a3d = \dfrac{{z \times M}}{{{N_A} \times {a^3}}} to calculate the edge length of the unit cell, which can also be written as:
a=z×MNA×d3a = \sqrt[3]{{\dfrac{{z \times M}}{{{N_A} \times d}}}}
Formula used:
-Density of a unit cell: d=z×MNA×a3d = \dfrac{{z \times M}}{{{N_A} \times {a^3}}} (A)
Where, d = density of the unit cell;
z = number of atoms present in the unit cell;
M = molar mass of the atom;
NA{N_A}= Avogadro number;
a = side length of the unit cell.

Complete answer:
-First of all we need to see what a body centred cubic crystal (bcc) is.
In a body centred cubic unit cell or crystal atoms are present at each of its 8 corners and one atom is present at the body centre.
The total number of atoms present in a bcc crystal are:
(1)8 corner atoms × (1/8) contribution of each corner atom = 1 atom
(2)1 body centre atom
So, in total 2 atoms are present in a bcc crystal. It looks like:

-The values for a Mo bcc structure given in the question are: d = 10.3g.cm3g.c{m^{ - 3}}, M = 95.94g.mol1g.mo{l^{ - 1}} and we know that NA{N_A}= 6.023×10236.023 \times {10^{23}}. Also in the above discussion we just saw that for a bcc crystal z = 2.
Since we have d, M, NA{N_A}and z; we can find out the value of ‘a’ using the equation (A).
d=z×MNA×a3d = \dfrac{{z \times M}}{{{N_A} \times {a^3}}}
10.3=2×95.946.023×1023×a310.3 = \dfrac{{2 \times 95.94}}{{6.023 \times {{10}^{23}} \times {a^3}}}
a3=2×95.946.023×1023×10.3{a^3} = \dfrac{{2 \times 95.94}}{{6.023 \times {{10}^{23}} \times 10.3}}
= 191.8862.0369×1023\dfrac{{191.88}}{{62.0369 \times {{10}^{23}}}}
= 3.092997×1023cm33.092997 \times {10^{ - 23}}c{m^3}
a=3.092997×10233a = \sqrt[3]{{3.092997 \times {{10}^{ - 23}}}}
= 3.14×108cm3.14 \times {10^{ - 8}}cm
= 3.14 Å
So, the side of the cubic crystal is 3.14 Å.

Note:
In a bcc crystal or lattice there are only 2 atoms present and so it has a packing efficiency of 68%. Iron, chromium, tungsten and niobium also exhibit bcc structure.