Question:
An element (atomic mass $100 \,g /\, mol$ ) having $BCC$ structure has unit cell edge $400\, pm$. The density of element is (No. of atom in $BCC ( Z )=2$ ).
An element (atomic mass $100 \,g /\, mol$ ) having $BCC$ structure has unit cell edge $400\, pm$. The density of element is (No. of atom in $BCC ( Z )=2$ ).
Updated On: Apr 28, 2024
- $10.376\, g/cm^3$
- $5.1888 \,g/cm^3$
- $7.289\, g/cm^3$
- $2.144 \,g/cm^3$
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The Correct Option is B
Solution and Explanation
No. of atoms per unit cell in $b.c.c.$ lattice $(n) = 2$
Density, $d = \frac{n \times M}{a^3 \times N_A}$
$ = \frac{2\times 100}{(4\times 10^{-8}\,cm)^3 \times 6.02 \times 10^{23}}$
$ = \frac{200}{38.528} $
$ = 5.19\,g/cm^3$.
Density, $d = \frac{n \times M}{a^3 \times N_A}$
$ = \frac{2\times 100}{(4\times 10^{-8}\,cm)^3 \times 6.02 \times 10^{23}}$
$ = \frac{200}{38.528} $
$ = 5.19\,g/cm^3$.
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