Question

Calculate the number of unit cells in \(1\text{cm}^3\) of metal if it forms simple cubic structure with unit cell edge length 500 pm.

• A. \(4.0\times 10^{21}\)
• B. \(2.0\times 10^{21}\)
• C. \(8.0\times 10^{21}\)
• D. \(3.0\times 10^{21}\)

Hint:

Divide 1 cm\(^3\) by the volume of one cell, \(a^3\) with \(a = 500\) pm \(=5\times10^{-8}\) cm.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The number of unit cells equals total volume divided by the volume of one unit cell. For a cube of edge \(a\), the cell volume is \(a^3\).

Step 2: Detailed Explanation
Edge: \(500\ \text{pm} = 5\times10^{-8}\) cm.
\[ a^3 = (5\times10^{-8})^3 = 125\times10^{-24} = 1.25\times10^{-22}\ \text{cm}^3 \]
\[ N = \frac{1}{1.25\times10^{-22}} = 8.0\times10^{21} \]
The other options come from a wrong cube or a wrong conversion of picometres.

Final Answer:
There are \(8.0\times10^{21}\) unit cells in 1 cm\(^3\), option (C). \[ \boxed{8.0\times10^{21}\ \text{(C)}} \]