The packing efficiency of the two-dimensional | vedclass

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(D) In the given two-dimensional square unit cell,there are four corner atoms and one central atom. Each corner atom contributes $1/4$ of its area to

Vedclass · Accepted Answer

$78.54$

Vedclass · Answer

(D) In the given two-dimensional square unit cell,there are four corner atoms and one central atom. Each corner atom contributes $1/4$ of its area to the unit cell,so $4 	imes (1/4) = 1$ atom from the corners. The central atom contributes $1$ full atom to the unit cell. Total number of atoms $= 1 + 1 = 2$. Let the radius of the atom be $r$ and the side of the square be $a$. The diagonal of the square is $a\sqrt{2} = 4r$,so $a = 4r/\sqrt{2} = 2r\sqrt{2}$. Packing efficiency $= \frac{	ext{Area of 2 atoms}}{	ext{Area of square}} 	imes 100 = \frac{2 	imes \pi r^2}{a^2} 	imes 100$. Substituting $a = 2r\sqrt{2}$,we get: Packing efficiency $= \frac{2 \pi r^2}{(2r\sqrt{2})^2} 	imes 100 = \frac{2 \pi r^2}{8r^2} 	imes 100 = \frac{\pi}{4} 	imes 100 \approx 78.54 \%$.

List-$I$	List-$II$
$(i) \ a \bar{a}$	$(A) -\frac{\pi}{3}$
$(ii) \ \arg \left(\frac{1}{\bar{a}}\right)$	$(B) -i \sqrt{3}$
$(iii) \ a-\bar{a}$	$(C) \frac{2i}{\sqrt{3}}$
$(iv) \ \operatorname{Im}\left(\frac{4}{3a}\right)$	$(D) 1$
	$(E) \frac{\pi}{3}$
	$(F) \frac{2}{\sqrt{3}}$

The packing efficiency of the two-dimensional square unit cell shown below is ........... $ \% $

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