GATE 2023 ME – Question 51
The atomic radius of a hypothetical face-centered cubic (FCC) metal is $(\sqrt{2}/10)$ nm. The atomic weight of the metal is 24.092 g/mol. Taking Avogadro's number to be $6.023 \times 10^{23}$ atoms/mol, the density of the metal is ____________ kg/m$^3$.
(Answer in integer)
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Correct answer: 2500
Explanation
For an FCC lattice the atoms touch along the face diagonal, $4r = \sqrt{2}a$, so $a = 2\sqrt{2}r = 2\sqrt{2} \times \frac{\sqrt{2}}{10} = 0.4$ nm $= 4 \times 10^{-10}$ m. The cell volume is $a^3 = 6.4 \times 10^{-29}$ m$^3$ and it holds 4 atoms. The mass of the 4 atoms is $\frac{4 \times 24.092 \times 10^{-3}}{6.023 \times 10^{23}} = 1.6 \times 10^{-25}$ kg. The density is $\frac{1.6 \times 10^{-25}}{6.4 \times 10^{-29}} = 2500$ kg/m$^3$.