Question:
A metal crystallizes in simple cubic lattice. The radius of the metal atom is $x\, pm$. What is the volume of the unit cell in $pm^3$?
A metal crystallizes in simple cubic lattice. The radius of the metal atom is $x\, pm$. What is the volume of the unit cell in $pm^3$?
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In simple cubic structure, unit cell edge length is twice the atomic radius, so volume is cube of edge length.
Updated On: Jan 5, 2026
- \(x^3\)
- \(4x^3\)
- \(8x^3\)
- \(16x^3\)
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The Correct Option is C
Solution and Explanation
In simple cubic lattice: \[ \text{Edge length} = a = 2r \] Given radius \(r = x\, pm\), so \[ a = 2x \] Volume of unit cell: \[ V = a^3 = (2x)^3 = 8x^3\, pm^3 \]
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