GATE 2026 ME – Question 62
A metal has FCC crystal structure with 2.71 g/cm$^3$ of density and 26.98 g/mol of atomic weight. The Avogadro's number is $6.023 \times 10^{23}$. The atomic radius of the metal is ________ nm (*rounded off to 2 decimal places*).
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Correct answer: 0.13 to 0.15
Explanation
An FCC unit cell holds 4 atoms, so its volume is $a^3 = \frac{4 \times 26.98}{2.71 \times 6.023 \times 10^{23}} = 6.61 \times 10^{-23}$ cm$^3$. This gives $a = 4.04 \times 10^{-8}$ cm, which is 0.404 nm. In an FCC lattice the atoms touch along a face diagonal, so $r = \frac{a}{2\sqrt{2}} = \frac{0.404}{2.828} = 0.14$ nm.