Question

A metal crystallises in face centred cubic structure with metallic radius 2​A∘. The volume of the unit cell (in m³) is

• A. 6.4 x 10^-30
• B. 4 x 10^-10
• C. 6.4 x 10^-29
• D. 4 x 10^-9

Answer 1

The Correct Option is C

In a face-centered cubic (FCC) structure, the relation between the edge length \( a \) of the unit cell and the atomic radius \( r \) is given by: \[ a = \frac{4r}{\sqrt{2}} \] Here, the metallic radius \( r = \sqrt{2} \, \text{Å} \), so: \[ a = \frac{4 \times \sqrt{2}}{\sqrt{2}} = 4 \, \text{Å} \] Now, the volume of the unit cell \( V \) is given by: \[ V = a^3 \] Substitute \( a = 4 \, \text{Å} \): \[ V = (4 \times 10^{-10} \, \text{m})^3 = 6.4 \times 10^{-29} \, \text{m}^3 \] Thus, the volume of the unit cell is \( 6.4 \times 10^{-29} \, \text{m}^3 \).

The correct option is (C) : \( 6.4 \times 10^{-29}\)

Answer 2

In a face-centered cubic (FCC) structure, the relation between the metallic radius \( r \) and the edge length \( a \) of the unit cell is given by: \[ a = \frac{4r}{\sqrt{2}} \] The volume \( V \) of the unit cell is given by: \[ V = a^3 \] Substitute the expression for \( a \): \[ V = \left( \frac{4r}{\sqrt{2}} \right)^3 = \frac{64r^3}{2\sqrt{2}} = \frac{64r^3}{2\sqrt{2}} \] Given that the metallic radius is \( r = \sqrt{2} \, \text{Å} = \sqrt{2} \times 10^{-10} \, \text{m} \), we substitute this value into the equation. So, calculating the value of \( V \): \[ V = 6.4 \times 10^{-29} \, \text{m}^3 \]

Thus, the correct answer is \( {(C)} \), and the volume of the unit cell is \( 6.4 \times 10^{-29} \, \text{m}^3 \).