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Aluminium crystallizes in a cubic close ...

Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm. The edge length of the unit cell and number of unit cells per cc of aluminium respectively are

A

354 pm , `2.25 xx 10^22`

B

648 pm , ` 3.3 xx 10^23`

C

298 pm ,`1.25 xx 10^22`

D

436 pm , `6.3 xx 10^22`

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To solve the problem of finding the edge length of the unit cell and the number of unit cells per cc of aluminum, we can follow these steps: ### Step 1: Determine the relationship between the edge length and the metallic radius. Aluminum crystallizes in a face-centered cubic (FCC) structure. In an FCC unit cell, the face diagonal is equal to four times the radius of the atom (4r), and the face diagonal can also be expressed in terms of the edge length (a) of the cube using the Pythagorean theorem. The face diagonal (d) can be calculated as: \[ d = a\sqrt{2} \] Setting the two expressions for the face diagonal equal gives us: \[ a\sqrt{2} = 4r \] ### Step 2: Substitute the given metallic radius into the equation. Given that the metallic radius (r) of aluminum is 125 pm (picometers), we can substitute this value into the equation: \[ a\sqrt{2} = 4 \times 125 \, \text{pm} \] \[ a\sqrt{2} = 500 \, \text{pm} \] ### Step 3: Solve for the edge length (a). Now, we can solve for the edge length (a): \[ a = \frac{500 \, \text{pm}}{\sqrt{2}} \] \[ a = \frac{500}{1.414} \, \text{pm} \] \[ a \approx 353.55 \, \text{pm} \] ### Step 4: Convert the edge length from picometers to centimeters. To convert the edge length from picometers to centimeters, we use the conversion factor \(1 \, \text{pm} = 10^{-12} \, \text{m} = 10^{-10} \, \text{cm}\): \[ a \approx 353.55 \, \text{pm} \times 10^{-10} \, \text{cm/pm} \] \[ a \approx 3.5355 \times 10^{-8} \, \text{cm} \] ### Step 5: Calculate the volume of the unit cell. The volume (V) of the cubic unit cell can be calculated using the formula: \[ V = a^3 \] Substituting the value of a: \[ V = (3.5355 \times 10^{-8} \, \text{cm})^3 \] \[ V \approx 4.42 \times 10^{-24} \, \text{cm}^3 \] ### Step 6: Calculate the number of unit cells per cubic centimeter. To find the number of unit cells per cubic centimeter, we take the reciprocal of the volume of one unit cell: \[ \text{Number of unit cells per cm}^3 = \frac{1}{V} \] \[ \text{Number of unit cells per cm}^3 = \frac{1}{4.42 \times 10^{-24} \, \text{cm}^3} \] \[ \text{Number of unit cells per cm}^3 \approx 2.26 \times 10^{23} \] ### Final Answers: - Edge length of the unit cell: **353.55 pm** (or **3.54 x 10^-8 cm**) - Number of unit cells per cc of aluminum: **2.26 x 10^23**

To solve the problem of finding the edge length of the unit cell and the number of unit cells per cc of aluminum, we can follow these steps: ### Step 1: Determine the relationship between the edge length and the metallic radius. Aluminum crystallizes in a face-centered cubic (FCC) structure. In an FCC unit cell, the face diagonal is equal to four times the radius of the atom (4r), and the face diagonal can also be expressed in terms of the edge length (a) of the cube using the Pythagorean theorem. The face diagonal (d) can be calculated as: \[ d = a\sqrt{2} \] ...
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