Question

An element having a face-centered cubic structure has a density of $6.23gc{{m}^{-3}}$ . The atomic mass of elements is 60. The edge length of the unit cell is:
A. 400pm
B. 300pm
C. 425pm
D. 370pm

Answer 1

Hint As we know that FCC is called a face centred cubic unit cell. It is found that FCC contains four atoms per unit cell and has a coordination number of 12. In order to calculate the edge length of unit cell we will use the formula:
${{a}^{3}}=\dfrac{z\times M}{{{N}_{A}}\times d}$

Complete Step by step solution:
- As we are being provided with the information that the element having a face-centered cubic structure has a density of $6.23gc{{m}^{-3}}$ . The atomic mass of elements is 60. Now, we will calculate the edge length of unit cell we will use the formula:
${{a}^{3}}=\dfrac{z\times M}{{{N}_{A}}\times d}$
- Where,
a = edge length
z = total number of atoms
M = atomic mass of the element
d = density
${{N}_{A}}$= Avogadro’s number
- Now, by putting all the values given in the above equation we get:
${{a}^{3}}=\dfrac{4\times 60}{6.023\times {{10}^{23}}\times 6.23}$
- Now, by solving we get:
$\begin{align}
& {{a}^{3}}=64\times {{10}^{-24}}c{{m}^{3}} \\
& \Rightarrow a=\left( 4\times {{10}^{-8}}\times {{10}^{10}} \right)pm=400pm \\
\end{align}$

- Hence, we can conclude that the edge length of the unit cell is 400 pm.

Note:
- We must not forget to write the unit after solving any question.
- We must not get confused in terms of FCC and BCC. FCC is face centred cubic unit cell, it contains four atoms per unit cell. Whereas, BCC is a body centred unit cell, it contains 2 atoms per unit cell.