Question:
An elemental crystal has a density of $8570 \,kg / m ^{3}$. The packing efficiency is $0.68$. The closest distance of approach between neighbouring atom is $2.86 \,?$. What is the mass of one atom approximately?
An elemental crystal has a density of $8570 \,kg / m ^{3}$. The packing efficiency is $0.68$. The closest distance of approach between neighbouring atom is $2.86 \,?$. What is the mass of one atom approximately?
Updated On: Dec 16, 2025
- 93 amu
- 39 amu
- 63 amu
- 29 amu
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The Correct Option is A
Solution and Explanation
The packing efficiency $= 0.68$, means the given lattice is BCC.
The closest distance of approach $= 2r$
$2r = 2.86\, ? = a \sqrt{3}$
or $a = \frac{ 2 \times 2.86}{\sqrt{3}} = 3.30\,? $
Let atomic weight of the element $= a$
$\therefore \frac{2 \times9}{36 \times10^{23} \times\left(3.3\right)^{3} \times10^{-24}} = 8.57 $
$a = 8.57 \times 3 \times \left(3.3\right)^{3} \times 0.1$
$ = 92.39 \simeq 93 $ amu
The closest distance of approach $= 2r$
$2r = 2.86\, ? = a \sqrt{3}$
or $a = \frac{ 2 \times 2.86}{\sqrt{3}} = 3.30\,? $
Let atomic weight of the element $= a$
$\therefore \frac{2 \times9}{36 \times10^{23} \times\left(3.3\right)^{3} \times10^{-24}} = 8.57 $
$a = 8.57 \times 3 \times \left(3.3\right)^{3} \times 0.1$
$ = 92.39 \simeq 93 $ amu
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