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An elemental crystal has a density of $8570 \mathrm{~kg} / \mathrm{m}^{3}$. The packing efficiency is 0.68 . The closest distance of approach between neighbouring atom is $2.86 Ã…$. What is the mass of one atom approximately?
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The correct answer is:
$93 \mathrm{amu}$
The packing efficiency $=0.68$, means the given lattice is $\mathrm{BCC}$. The closest distance of approach $=2 \mathrm{r}$
$$
\begin{array}{l}
2 r=2.86 A=a \sqrt{3} \\
\text { or a }=\frac{2 \times 2.86}{\sqrt{3}}=3.30 Ã…
\end{array}
$$
Let atomic weight of the element $=\mathrm{a}$
$$
\begin{aligned}
\therefore & \frac{2 \times 9}{36 \times 10^{23} \times(3.3)^{3} \times 10^{-24}}=8.57 \\
\mathrm{a} &=8.57 \times 3 \times(3.3)^{3} \times 0.1 \\
&=92.39 \simeq 93 \mathrm{amu}
\end{aligned}
$$
$$
\begin{array}{l}
2 r=2.86 A=a \sqrt{3} \\
\text { or a }=\frac{2 \times 2.86}{\sqrt{3}}=3.30 Ã…
\end{array}
$$
Let atomic weight of the element $=\mathrm{a}$
$$
\begin{aligned}
\therefore & \frac{2 \times 9}{36 \times 10^{23} \times(3.3)^{3} \times 10^{-24}}=8.57 \\
\mathrm{a} &=8.57 \times 3 \times(3.3)^{3} \times 0.1 \\
&=92.39 \simeq 93 \mathrm{amu}
\end{aligned}
$$
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