Question

# The element Chromium exists in a bcc lattice with unit cell edge $2.88 \times {10^{ - 10}}m$. The density of Chromium is $7.2 \times {10^3}Kg\;{m^{ - 3}}$. How many atoms does $52 \times {10^{ - 3}}kg$ of chromium contain?

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Hint: As we know that by knowing the dimensions of a unit cell, we can calculate the volume of the unit cell, similarly by knowing the density of a unit cell we can calculate the mass and number of atoms in the unit cell. We also know that the bcc structure of a unit cell possesses eight lattice points at eight corners and one additional lattice point at the centre of the body.

Now, we are given that the edge length of bcc lattice of chromium elements is $2.88 \times {10^{ - 10}}m$.
And the density of the chromium is given as $7.2 \times {10^3}Kg\;{m^{ - 3}}$.
Let us recall the formula to calculate the density of the given unit cell which is $d = \dfrac{{Z \times M}}{{{a^3} \times N}}$ where $N$, is the Avogadro ’s number is in general but here we need to calculate the number of atoms which will be given by the $N$corresponding to the molecular mass of the chromium. And mass of chromium is given $52 \times {10^{ - 3}}kg$.
And we know that for a body centred cubic unit cell the number of atoms per unit cell or the value of $Z$ is $2$.
$N = \dfrac{{Z \times M}}{{{a^3} \times d}}$
$\Rightarrow N = \dfrac{{2 \times 52 \times {{10}^{ - 3}}}}{{{{(2.88 \times {{10}^{ - 10}})}^3} \times 7.2 \times {{10}^3}}}$
$\Rightarrow N = 6.04 \times {10^{23}}atoms$
Therefore the correct answer is $6.04 \times {10^{23}}atoms$.
Note: Remember that the body centred cubic arrangement contains $8$ lattice point at the corners and each one of these is shared by $8$cells and another lattice point is completely inside the unit cell or present at the centre of the body thus the number of points or atoms per unit cell of bcc is $8 \times \dfrac{1}{8} + 1 = 2$, therefore the value of $Z = 2$. Similarly, the value of ccp arrangement can be calculated which comes out to be $Z = 4$.