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The interplanar spacing of (110) planes in a cubic unit cell with lattice parameter a = 4.242 Å is:
  • a)
  • b)
    63Å
  • c)
    7.35Å
  • d)
    2.45Å
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The interplanar spacing of (110) planes in a cubic unit cell with latt...
Given:
Lattice parameter, a = 4.242 Å
Miller indices of plane, (110)

To find:
Interplanar spacing of (110) planes

Solution:
The interplanar spacing (d) of a plane with Miller indices (hkl) can be calculated using the formula:

d = a / √(h² + k² + l²)

For the (110) plane, the Miller indices are (hkl) = (1 1 0)

Substituting the values in the formula, we get:

d = 4.242 / √(1² + 1² + 0²)
d = 4.242 / √2
d = 4.242 / 1.414
d = 3 Å

Therefore, the interplanar spacing of (110) planes in a cubic unit cell with lattice parameter a = 4.242 Å is 3 Å.

Answer: Option A (3)
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The interplanar spacing of (110) planes in a cubic unit cell with lattice parameter a = 4.242 Å is:a)3Åb)63Åc)7.35Åd)2.45ÅCorrect answer is option 'A'. Can you explain this answer?
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