Question

16. The density of a fcc element(atomic mass 60.2) is 6.25 g/cm3. Calculate the edge length of the unit cell *

Answer 1

Answer:

• The edge length (a) is 400 pm

Given:

1. Density (ρ) = 6.25 g/cm³
2. Atomic mass (M) = 60.2 g

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\large\bigstar\;\underline{\boxed{\sf \rho=\dfrac{z\times M}{N_{0}\times a^{3}}}}$

Here,

• z Denotes no. of atoms per unit cell.
• M Denotes atomic mass.
• N₀ Denotes Avogadro's number.
• a Denotes edge length.

We know,

For a FCC unit cell, no. of atoms per unit cell is 4.

Solving & Substituting the values,

$\displaystyle\longrightarrow \sf a^{3}=\dfrac{z\times M}{N_{0}\times \rho}\\\\\\\longrightarrow\sf a^{3}=\dfrac{4\times 60.2}{6.02\times 10^{23}\times 6.25}\\\\\\\longrightarrow\sf a^{3}=\dfrac{240.8}{37.625\times 10^{23}}\\\\\\\longrightarrow\sf a^{3}=\dfrac{6.4}{10^{23}}\\\\\\\longrightarrow\sf a^{3}=6.4\times 10^{-23}\\\\\\\longrightarrow\sf a^{3}=64\times 10^{-24}$

$\displaystyle\longrightarrow\sf a=\sqrt{64\times 10^{-24}}\\\\\\\longrightarrow\sf a=4\times 10^{-8}\;cm\\\\\\\longrightarrow\sf a=400\times 10^{-10}\;m\\\\\\\longrightarrow\sf \large{\underline{\boxed{\red{\sf a=400\;pm}}}}$

∴ The edge length (a) is 400 pm.

$\rule{300}{1.5}$