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Calculate the atomic radius of elementary silver which crystallises in face centred cubic lattice with unit cell edge length `4.086 xx 10^(-10) m`.

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To calculate the atomic radius of elementary silver which crystallizes in a face-centered cubic (FCC) lattice with a unit cell edge length of \( a = 4.086 \times 10^{-10} \) m, we can follow these steps: ### Step 1: Understand the FCC Structure In a face-centered cubic lattice, atoms are located at each corner of the cube and at the center of each face. The arrangement of the atoms allows us to derive a relationship between the atomic radius \( r \) and the edge length \( a \) of the unit cell. ### Step 2: Derive the Relationship For an FCC lattice, the relationship between the atomic radius \( r \) and the edge length \( a \) can be derived using the geometry of the cube. The diagonal of the face of the cube can be expressed in terms of \( a \) and \( r \): - The diagonal of the face (let's call it \( AC \)) can be calculated using Pythagoras' theorem: \[ AC = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \] - In terms of atomic radii, this diagonal \( AC \) is equal to four times the atomic radius \( r \) (since there are two radii from the corner atom to the face-centered atom): \[ AC = 4r \] ### Step 3: Set Up the Equation From the above relationships, we can equate the two expressions for \( AC \): \[ 4r = a\sqrt{2} \] ### Step 4: Solve for the Atomic Radius \( r \) Rearranging the equation to solve for \( r \): \[ r = \frac{a\sqrt{2}}{4} \] ### Step 5: Substitute the Value of \( a \) Now, substitute the given value of \( a \): \[ a = 4.086 \times 10^{-10} \text{ m} \] \[ r = \frac{4.086 \times 10^{-10} \times \sqrt{2}}{4} \] ### Step 6: Calculate \( r \) Calculating \( \sqrt{2} \approx 1.414 \): \[ r = \frac{4.086 \times 10^{-10} \times 1.414}{4} \] \[ r = \frac{5.775 \times 10^{-10}}{4} \] \[ r = 1.44375 \times 10^{-10} \text{ m} \] ### Final Answer The atomic radius of elementary silver is approximately: \[ r \approx 1.44 \times 10^{-10} \text{ m} \]

To calculate the atomic radius of elementary silver which crystallizes in a face-centered cubic (FCC) lattice with a unit cell edge length of \( a = 4.086 \times 10^{-10} \) m, we can follow these steps: ### Step 1: Understand the FCC Structure In a face-centered cubic lattice, atoms are located at each corner of the cube and at the center of each face. The arrangement of the atoms allows us to derive a relationship between the atomic radius \( r \) and the edge length \( a \) of the unit cell. ### Step 2: Derive the Relationship For an FCC lattice, the relationship between the atomic radius \( r \) and the edge length \( a \) can be derived using the geometry of the cube. The diagonal of the face of the cube can be expressed in terms of \( a \) and \( r \): ...
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