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In fcc crystal lattice, edge length is 400 pm. Find the diameter of greatest sphere which can be fitted into the interstitial void without distortion of lattice:
- a)158.54 pm
- b)117.08 pm
- c)129.54 pm
- d)414 pm
Correct answer is option 'B'. Can you explain this answer?
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In fcc crystal lattice, edge length is 400 pm. Find the diameter of gr...
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In fcc crystal lattice, edge length is 400 pm. Find the diameter of gr...
Given information:
- Crystal lattice: FCC (face-centered cubic)
- Edge length of the lattice: 400 pm
To find:
- Diameter of the greatest sphere that can be fitted into the interstitial void without distorting the lattice
Solution:
Step 1: Visualize the FCC lattice structure
- The face-centered cubic (FCC) lattice is a common arrangement of atoms in a crystal lattice.
- In an FCC lattice, atoms are arranged in a cubic unit cell with additional atoms located at the center of each face of the cube.
- This arrangement creates interstitial voids, which are empty spaces between atoms in the lattice.
Step 2: Determine the size of the interstitial void
- To find the diameter of the greatest sphere that can fit into the interstitial void, we need to determine the size of the void.
- In an FCC lattice, the interstitial void is octahedral in shape, which means it can be imagined as two back-to-back tetrahedra.
- The distance between the centers of the atoms in the FCC lattice is equal to the edge length of the unit cell, which is given as 400 pm.
Step 3: Calculate the size of the interstitial void
- The distance between the centers of the two atoms in the FCC lattice is equal to the edge length of the unit cell, which is 400 pm.
- The distance between the centers of the two atoms in the same face of the cube is equal to half the edge length, which is 200 pm.
- The distance between the centers of the two atoms in the same face plane and the atom in the center of the cube is equal to the diagonal of a face of the cube, which can be calculated using Pythagoras' theorem.
- The diagonal of a face of a cube with edge length 400 pm is √(200^2 + 200^2) = √(2 * 200^2) = 282.84 pm.
- Therefore, the size of the interstitial void is 282.84 pm.
Step 4: Calculate the diameter of the greatest sphere
- The diameter of the greatest sphere that can fit into the interstitial void is equal to the size of the void.
- Therefore, the diameter of the greatest sphere is 282.84 pm, which is approximately 117.08 pm.
Conclusion:
The correct answer is option 'B', 117.08 pm.
- Crystal lattice: FCC (face-centered cubic)
- Edge length of the lattice: 400 pm
To find:
- Diameter of the greatest sphere that can be fitted into the interstitial void without distorting the lattice
Solution:
Step 1: Visualize the FCC lattice structure
- The face-centered cubic (FCC) lattice is a common arrangement of atoms in a crystal lattice.
- In an FCC lattice, atoms are arranged in a cubic unit cell with additional atoms located at the center of each face of the cube.
- This arrangement creates interstitial voids, which are empty spaces between atoms in the lattice.
Step 2: Determine the size of the interstitial void
- To find the diameter of the greatest sphere that can fit into the interstitial void, we need to determine the size of the void.
- In an FCC lattice, the interstitial void is octahedral in shape, which means it can be imagined as two back-to-back tetrahedra.
- The distance between the centers of the atoms in the FCC lattice is equal to the edge length of the unit cell, which is given as 400 pm.
Step 3: Calculate the size of the interstitial void
- The distance between the centers of the two atoms in the FCC lattice is equal to the edge length of the unit cell, which is 400 pm.
- The distance between the centers of the two atoms in the same face of the cube is equal to half the edge length, which is 200 pm.
- The distance between the centers of the two atoms in the same face plane and the atom in the center of the cube is equal to the diagonal of a face of the cube, which can be calculated using Pythagoras' theorem.
- The diagonal of a face of a cube with edge length 400 pm is √(200^2 + 200^2) = √(2 * 200^2) = 282.84 pm.
- Therefore, the size of the interstitial void is 282.84 pm.
Step 4: Calculate the diameter of the greatest sphere
- The diameter of the greatest sphere that can fit into the interstitial void is equal to the size of the void.
- Therefore, the diameter of the greatest sphere is 282.84 pm, which is approximately 117.08 pm.
Conclusion:
The correct answer is option 'B', 117.08 pm.
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In fcc crystal lattice, edge length is 400 pm. Find the diameter of gr...
Ans.
For an octahedral void a = 2 (r + R) In fcc lattice the largest void present is octahedral void. If the radius of void sphere is R and of lattice sphere is r. Then,
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In fcc crystal lattice, edge length is 400 pm. Find the diameter of greatest sphere which can be fitted into the interstitial void without distortion of lattice:a)158.54 pmb)117.08 pmc)129.54 pmd)414 pmCorrect answer is option 'B'. Can you explain this answer?
Question Description
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Solutions for In fcc crystal lattice, edge length is 400 pm. Find the diameter of greatest sphere which can be fitted into the interstitial void without distortion of lattice:a)158.54 pmb)117.08 pmc)129.54 pmd)414 pmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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