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: Count cells along an edge
1 cm $=10^{10}$ pm, so along one edge there are $\dfrac{10^{10}}{500} = 2\times10^{7}$ cells.

Step 2: Cube it
Cells in the cube: $(2\times10^{7})^3 = 8\times10^{21}$.
So 1 cm$^3$ holds $8.0\times10^{21}$ unit cells, option (C).

Final Answer:
The count is $8.0\times10^{21}$ unit cells, option (C). \[ \boxed{8.0\times10^{21}} \]