Question

How many unit cells are present in 100 g of an element with fcc crystal structure having density \(10\ \text{g cm}^{-3}\) and edge length \(100\ \text{pm}\)?

• A. \(3 \times 10^{25}\)
• B. \(2 \times 10^{25}\)
• C. \(4 \times 10^{25}\)
• D. \(1 \times 10^{25}\)

Hint:

Number of unit cells \(=\) total volume of crystal \(/\) volume of one unit cell.

Answer 1

The Correct Option is D

Step 1: Convert edge length into cm.
\[ a = 100\ \text{pm} = 100 \times 10^{-10}\ \text{cm} = 1 \times 10^{-8}\ \text{cm} \]

Step 2: Calculate volume of one unit cell.
\[ V_{\text{cell}} = a^3 = (1 \times 10^{-8})^3 = 1 \times 10^{-24}\ \text{cm}^3 \]

Step 3: Find total volume of the element.
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{100}{10} = 10\ \text{cm}^3 \]

Step 4: Calculate number of unit cells.
\[ \text{Number of unit cells} = \frac{10}{1 \times 10^{-24}} = 1 \times 10^{25} \]

Step 5: Conclusion.
The number of unit cells present is \(1 \times 10^{25}\).