Question

A metal crystallizes in cubic lattice with edge length of \(4\ \text{\AA}\). The number of unit cells of this metal present in \(128\ \text{cm}^3\) volume is

• A. \(2\times10^{24}\)
• B. \(4\times10^{24}\)
• C. \(8\times10^{24}\)
• D. \(16\times10^{24}\)

Hint:

For a cubic crystal, \[ \boxed{ \text{Number of unit cells} = \frac{\text{Total volume}}{a^3} } \] where \(a\) is the edge length of the unit cell.

Answer 1

The Correct Option is A

Step 1: Convert the edge length into centimeters. Since, \[ 1\ \text{\AA}=10^{-8}\ \text{cm}, \] \[ a=4\times10^{-8}\ \text{cm}. \]

Step 2: Calculate the volume of one unit cell. \[ V_{\text{cell}} = a^3 = (4\times10^{-8})^3 = 64\times10^{-24} = 6.4\times10^{-23}\ \text{cm}^3. \]

Step 3: Calculate the number of unit cells. \[ N = \frac{128}{6.4\times10^{-23}} = 20\times10^{23} = 2\times10^{24}. \] Hence, \[ \boxed{2\times10^{24}}. \] Thus, \[ \boxed{(A)} \] is the correct answer.