A metal having atomic mass 60.23 g/mol cryst | vedclass

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Question

(C) The $ABCABC$ packing corresponds to a face-centered cubic $(FCC)$ structure.For an $FCC$ unit cell,the relation between edge length $(a)$ and atom

Vedclass · Accepted Answer

$0.54$

Vedclass · Answer

(C) The $ABCABC$ packing corresponds to a face-centered cubic $(FCC)$ structure.For an $FCC$ unit cell,the relation between edge length $(a)$ and atomic radius $(r)$ is $a = 2\sqrt{2}r$,so $r = \frac{a}{2\sqrt{2}}$.Given $a = 10 \ \mathring{A} = 10 	imes 10^{-8} \ cm$.Radius $r = \frac{10 	imes 10^{-8}}{2\sqrt{2}} \ cm$.Volume of a single metal atom $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (\frac{10 	imes 10^{-8}}{2\sqrt{2}})^3 \ cm^3$.Mass of a single metal atom $m = \frac{	ext{Atomic mass}}{N_A} = \frac{60.23}{6.023 	imes 10^{23}} = 10^{-22} \ g$.Density of the atom $ho = \frac{m}{V} = \frac{10^{-22}}{\frac{4}{3} \pi (\frac{10 	imes 10^{-8}}{2\sqrt{2}})^3} \approx 0.54 \ g/cm^3$.

$A$ metal having atomic mass $60.23 \ g/mol$ crystallises in $ABCABC$ close packing. Calculate the density of a single metal atom if the edge length of the unit cell is $10 \ \mathring{A}$. [Given: $N_A = 6.023 \times 10^{23}$]

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