Question

An element with density $6.8gc{m^{ - 3}}$, occurs in a bcc structure with cell edge length equal to $290pm$ . Calculate the number of atoms present in $200g$ of the element.
A. $2.4 \times {10^{42}}$
B. $1.2 \times {10^{42}}$
C. $1.2 \times {10^{24}}$
D. $2.4 \times {10^{24}}$

Answer 1

Hint: A bcc unit cell has atoms at each corner of the cube and one atom at the centre of the cube. The centre atom is wholly belonging to the unit cell in which it is present. So the total number of atoms in a bcc cell is two. One at the centre and one from the corners.

Complete Step by step answer: In a crystal, the smallest repeating unit of the lattice is unit cell. It is the building block of that crystal. The unit cell can be primitive cubic, body-centered cubic or face- centered cubic.
In primitive cubic unit cells, the atoms are present at the corners.
In face- centered cubic unit cell, the atoms are at all corners and at the centre of all faces of the cube
In body- centered cubic unit cell, the atoms are present all the corners of the cube as well as one atom is present at the centre of the cube which wholly belong to that unit cell. The number of atoms in bcc cell- 8 atoms at the corner $6 \times \dfrac{1}{2} = 3\;atoms$ $ \times \dfrac{1}{8}$ per corner atom = 1 atom and 1 body centre atom= 1 atom. That means the total number of atoms per unit cell is 2.
For density calculation, we have the formula as - $d = \dfrac{{n \times m}}{{{a_3} \times {N_A}}}$
Given in the question,
Edge length (a) = $290pm$
Density (d) = $6.8gc{m^{ - 3}}$
And for bcc n=2
NA = $6.022 \times {10^{23}}$
put all the values in the given formula we will get
$6.8 = \dfrac{{2 \times m}}{{{{(290 \times {{10}^{ - 10}})}^3} \times 6.022 \times {{10}^{23}}}}$
So we have $m = \dfrac{{6.8 \times 6.023 \times {{10}^{23}} \times 24.4 \times {{10}^{ - 30}}}}{2}$
On calculating we get $m = 50gmo{l^{ - 1}}$
Now we have to calculate total number of atom in $200g$ of this crystal,
 So total number of atoms can be calculate as =$\dfrac{{200g \times 6.023 \times {{10}^{23}}atoms/mol}}{{50gmo{l^{ - 1}}}} = 2.4 \times {10^{24}}\;atoms$

Therefore the correct option will be D.

Note: In primitive unit cell, there are 8 atoms at the corner of the unit cell, therefore, the total number of atoms in one unit cell will be $8 \times \dfrac{1}{8} = 1atom$. Similarly for face- centered cubic unit cell total number of atoms in one unit cell will be $8 \times \dfrac{1}{8} = 1atom$+ $6 \times \dfrac{1}{2} = 3atoms$= 4 atoms.