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The radius of an atom of an element is 500 pm, if it crystallizes as a face-centred cubic lattice, then the edge length of the unit cell is
- a)500 pm
- b)707 pm
- c)1000 pm
- d)1414 pm
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The radius of an atom of an element is 500 pm, if it crystallizes as a...
Explanation:
Given, the radius of an atom of an element is 500 pm.
Let's assume that the atom is placed at the corner of the unit cell.
In a face-centered cubic lattice, there are atoms at each of the corners and at the center of each face of the cube.
Therefore, the effective diameter of the unit cell can be calculated as follows:
The atom at the corner is in contact with the atoms at the center of two adjacent faces.
Therefore, the effective diameter of the unit cell can be given as:
Effective diameter = 2 × radius of the atom + diagonal of the face of the unit cell
Diagonal of the face of the unit cell can be calculated using Pythagoras theorem as follows:
Diagonal of the face of the unit cell = √2 × edge length of the unit cell
Therefore,
Effective diameter = 2 × 500 pm + √2 × edge length of the unit cell
Edge length of the unit cell can be calculated by rearranging the above equation as follows:
Edge length of the unit cell = (Effective diameter - 2 × radius of the atom) / √2
Substituting the given values in the above equation, we get
Edge length of the unit cell = (2 × 500 pm - 2 × 500 pm) / √2
Edge length of the unit cell = 1000 / √2 pm
Edge length of the unit cell = 707 pm (approx)
Hence, the edge length of the unit cell is 1414 pm (approx).
Given, the radius of an atom of an element is 500 pm.
Let's assume that the atom is placed at the corner of the unit cell.
In a face-centered cubic lattice, there are atoms at each of the corners and at the center of each face of the cube.
Therefore, the effective diameter of the unit cell can be calculated as follows:
The atom at the corner is in contact with the atoms at the center of two adjacent faces.
Therefore, the effective diameter of the unit cell can be given as:
Effective diameter = 2 × radius of the atom + diagonal of the face of the unit cell
Diagonal of the face of the unit cell can be calculated using Pythagoras theorem as follows:
Diagonal of the face of the unit cell = √2 × edge length of the unit cell
Therefore,
Effective diameter = 2 × 500 pm + √2 × edge length of the unit cell
Edge length of the unit cell can be calculated by rearranging the above equation as follows:
Edge length of the unit cell = (Effective diameter - 2 × radius of the atom) / √2
Substituting the given values in the above equation, we get
Edge length of the unit cell = (2 × 500 pm - 2 × 500 pm) / √2
Edge length of the unit cell = 1000 / √2 pm
Edge length of the unit cell = 707 pm (approx)
Hence, the edge length of the unit cell is 1414 pm (approx).
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Community Answer
The radius of an atom of an element is 500 pm, if it crystallizes as a...
Formula for edge length of FCC structure where radius is 'r' = 4r/√2
Hence, edge length = 4×500/√2
→2000/√2 = 1414 (approx.)
Hence, edge length = 4×500/√2
→2000/√2 = 1414 (approx.)
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The radius of an atom of an element is 500 pm, if it crystallizes as a face-centred cubic lattice, then the edge length of the unit cell isa)500 pmb)707 pmc)1000 pmd)1414 pmCorrect answer is option 'D'. Can you explain this answer?
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