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The distance between two successive (110) planes in a simple cubic lattice with lattice parameter ‘a’ is:
- a)√2a
- b)√3a
- c)2√2a
- d)a/√2
Correct answer is option 'D'. Can you explain this answer?
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The distance between two successive (110) planes in a simple cubic lattice can be found using the following formula:
d = a / sqrt(2)
where d is the distance between two successive (110) planes, a is the lattice parameter.
In a simple cubic lattice, the lattice parameter is equal to the edge length of the unit cell, so we have:
a = 2r
where r is the radius of the atoms occupying the lattice points.
Therefore, the distance between two successive (110) planes is:
d = a / sqrt(2) = (2r) / sqrt(2) = r * sqrt(2)
This means that the distance between two successive (110) planes is equal to the radius of the atoms multiplied by the square root of 2.
d = a / sqrt(2)
where d is the distance between two successive (110) planes, a is the lattice parameter.
In a simple cubic lattice, the lattice parameter is equal to the edge length of the unit cell, so we have:
a = 2r
where r is the radius of the atoms occupying the lattice points.
Therefore, the distance between two successive (110) planes is:
d = a / sqrt(2) = (2r) / sqrt(2) = r * sqrt(2)
This means that the distance between two successive (110) planes is equal to the radius of the atoms multiplied by the square root of 2.
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