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If the radius of an atom of an element which forms a body centered cubic unit cell is 173.2 $\mathrm{pm}$, the volume of unit cell in $\mathrm{cm}^3$ is
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Verified Answer
The correct answer is:
$6.4 \times 10^{-23}$
Given,
Radius of an atom in body centered cubic (bcc) unit cell $=173.2 \mathrm{pm}$.
$\because$ For bcc structure
$$
\sqrt{3} \cdot a=4 r
$$
where, $a=$ edge-length
$r=$ radius of atom
and $a^3=V$ (volume of cubic unit cell).
$\therefore \quad a=\frac{4}{\sqrt{3}} \cdot r$
or, $\quad a=\frac{4}{1.73} \times 173.2 \times 10^{-10} \mathrm{~cm}$
$a=400 \times 10^{-10} \mathrm{~cm}$
Therefore,
$$
\begin{aligned}
a^3 & =V=\left[400 \times 10^{-10}\right]^3 \\
& =6,40,00,000 \times 10^{-30} \\
& =6.4 \times 10^{-23} \mathrm{~cm}^3
\end{aligned}
$$
Hence, $6.4 \times 10^{-23} \mathrm{~cm}^3$ is the correct answer.
Radius of an atom in body centered cubic (bcc) unit cell $=173.2 \mathrm{pm}$.
$\because$ For bcc structure
$$
\sqrt{3} \cdot a=4 r
$$
where, $a=$ edge-length
$r=$ radius of atom
and $a^3=V$ (volume of cubic unit cell).
$\therefore \quad a=\frac{4}{\sqrt{3}} \cdot r$
or, $\quad a=\frac{4}{1.73} \times 173.2 \times 10^{-10} \mathrm{~cm}$
$a=400 \times 10^{-10} \mathrm{~cm}$
Therefore,
$$
\begin{aligned}
a^3 & =V=\left[400 \times 10^{-10}\right]^3 \\
& =6,40,00,000 \times 10^{-30} \\
& =6.4 \times 10^{-23} \mathrm{~cm}^3
\end{aligned}
$$
Hence, $6.4 \times 10^{-23} \mathrm{~cm}^3$ is the correct answer.
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