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Al (atomic mass =27) crystallises in a c...

Al (atomic mass =27) crystallises in a cubic system with edge length (a) equal to `4Å` its density is `2.7 g//cm^(3)` The type of the unit cell is:

A

Simple

B

Face centred

C

Body centred

D

None of these

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To determine the type of unit cell for aluminum (Al) given its atomic mass, edge length, and density, we can follow these steps: ### Step 1: Gather Given Information - Atomic Mass (M) of Al = 27 g/mol - Edge Length (a) = 4 Å = \(4 \times 10^{-8}\) cm - Density (D) = 2.7 g/cm³ ### Step 2: Convert Edge Length to Centimeters Since the density is given in g/cm³, we need to convert the edge length from angstroms to centimeters: \[ a = 4 \text{ Å} = 4 \times 10^{-8} \text{ cm} \] ### Step 3: Use the Density Formula The formula for density (D) in terms of the number of atoms per unit cell (Z), atomic mass (M), edge length (a), and Avogadro's number (N_A) is given by: \[ D = \frac{Z \times M}{a^3 \times N_A} \] Where: - \(N_A = 6 \times 10^{23} \text{ atoms/mol}\) ### Step 4: Rearranging the Formula to Find Z We can rearrange the formula to solve for Z: \[ Z = \frac{D \times a^3 \times N_A}{M} \] ### Step 5: Substitute the Values Now we substitute the known values into the equation: \[ Z = \frac{2.7 \text{ g/cm}^3 \times (4 \times 10^{-8} \text{ cm})^3 \times 6 \times 10^{23} \text{ atoms/mol}}{27 \text{ g/mol}} \] ### Step 6: Calculate \(a^3\) First, calculate \(a^3\): \[ a^3 = (4 \times 10^{-8})^3 = 64 \times 10^{-24} \text{ cm}^3 = 6.4 \times 10^{-23} \text{ cm}^3 \] ### Step 7: Substitute \(a^3\) Back into Z Now substitute \(a^3\) back into the equation for Z: \[ Z = \frac{2.7 \times 6.4 \times 10^{-23} \times 6 \times 10^{23}}{27} \] ### Step 8: Calculate Z Calculating the numerator: \[ 2.7 \times 6.4 \times 6 = 103.68 \] Now, substituting this back: \[ Z = \frac{103.68}{27} \approx 3.84 \] ### Step 9: Determine the Type of Unit Cell Since Z is approximately 4, this indicates that aluminum crystallizes in a face-centered cubic (FCC) unit cell, where Z = 4. ### Conclusion Thus, the type of unit cell for aluminum is: **Face-Centered Cubic (FCC)**

To determine the type of unit cell for aluminum (Al) given its atomic mass, edge length, and density, we can follow these steps: ### Step 1: Gather Given Information - Atomic Mass (M) of Al = 27 g/mol - Edge Length (a) = 4 Å = \(4 \times 10^{-8}\) cm - Density (D) = 2.7 g/cm³ ### Step 2: Convert Edge Length to Centimeters ...
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