Question

Calculate the number of unit cells in $1~cm^{3}$ of an element if unit cell edge length is $2.0\times10^{-8}$ cm.

• A. $3.78\times10^{23}$
• B. $2.61\times10^{23}$
• C. $1.25\times10^{23}$
• D. $4.61\times10^{23}$

Hint:

Volume of cube = $a^3$. Ensure units are consistent ($cm^3$ and $cm$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find how many unit cell volumes (\( a^3 \)) are contained in a total volume of \( 1 \text{ cm}^3 \).

Step 2: Key Formula or Approach:
\[ \text{Number of unit cells} = \frac{\text{Total Volume}}{\text{Volume of one unit cell}} = \frac{V}{a^3} \]

Step 3: Detailed Explanation:
Given: \( a = 2.0 \times 10^{-8} \text{ cm} \); Total Volume \( = 1 \text{ cm}^3 \).
Volume of one unit cell \( = (2.0 \times 10^{-8} \text{ cm})^3 = 8.0 \times 10^{-24} \text{ cm}^3 \).
Number of unit cells \( = \frac{1}{8.0 \times 10^{-24}} \).
\[ = \frac{10^{24}}{8} = 0.125 \times 10^{24} = 1.25 \times 10^{23} . \]

Step 4: Final Answer:
The number of unit cells is \( 1.25 \times 10^{23} \).