Question

Potassium crystallizes in a body-centered cubic unit cell. The mass of the unit cell is:
\[
  (A)1.29 \times {10^{ - 23}}gm \\
  (B)1.295 \times {10^{ - 22}}gm \\
  (C)6.2 \times {10^{ - 23}}gm \\
  (D)1.29 \times {10^{ - 24}}gm \\
 \]

Answer 1

Hint: A body-centered cubic unit cell also known as BCC unit cell is a type of cell that contains atoms at every corner of the cube and one atom at the centre of the cube. The number of atoms in a BCC unit cell is $ 2 $ .

Complete answer:
In the question, we are given that potassium $ \left( K \right) $ crystallizes in a BCC unit cell so we need to find the mass of the unit cell. First we should know that the total number of atoms present in the body-centered cubic lattice is $ 2 $ .
The number of atoms can be calculated in the following manner:
-Total number of atoms present at the corners of the cube $ = 8 $
-Fraction of each atom contributing in the unit cell or inside the boundaries of the cube $ = \dfrac{1}{8} $
-Number of body atoms $ = 1 $
Therefore $ 8 \times \dfrac{1}{8} + 1 = 2 $
Now let us calculate the mass of the unit cell.
We know that the atomic mass/weight of K is $ 39gm $ and also there are $ 6.022 \times {10^{23}} $ atoms present in one mole of any substance.
If $ 6.022 \times {10^{23}} $ atoms of potassium weigh $ 39gm $
 $ 1 $ atom will weigh $ = \dfrac{{39}}{{6.022 \times {{10}^{23}}}} $
And so $ 2 $ atoms will weigh $ = \dfrac{{2 \times 39}}{{6.022 \times {{10}^{23}}}} $ ( $ 2 $ atoms because it is given in the question that potassium is a BCC unit cell)
 $ = \dfrac{{78}}{{6.022 \times {{10}^{23}}}} $
 $ = 12.95 \times {10^{ - 23}}gm $
 $ = 1.295 \times {10^{ - 22}}gm $
Therefore the mass of the unit cell is $ = 1.295 \times {10^{ - 22}}gm $. Hence the correct option is B.

Note:
There are many different types of unit cells, for example cubic cell, face-centered, etc. the net number of atoms in every unit cell is different: simple cube $ = 1 $ atom per unit cell, FCC $ = 4 $ atoms per unit cell. And every element crystallizes in some of the other unit cells and the mass of the unit cell can be found in the same manner done above.