The number of atoms in 100 g of an fcc crystal with density $d = 10{\rm{g/c}}{{\rm{m}}^{\rm{3}}}$ and cell edge equal to 100 pm, is equal to

JEE CHEMISTRY MCQ: SOLID STATE

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The number of atoms in 100 g of an fcc crystal with density

d = 10\mathrm{g/c}{\mathrm{m}^\mathrm{3}}

and cell edge equal to 100 pm, is equal to

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Correct option: D

4 \times {10^{25}}

Option D (

4 \times {10^{25}}

) is correct because the number of atoms is calculated using the density, cell edge, Avogadro's number, and the number of atoms per unit cell in an FCC crystal.

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