If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters

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Roosevelt Houghton
Given
We are given the radius of lead atom R= 0.175 nm. We are asked to calculate the volume of the unit cell crystal structure of the lead $$\displaystyle{V}_{{c}}$$ in cubic meters.
Solution
To calculate the volume of the unit cell we should know the crystal structure unit cell shape, where lead has FCC crystal structure (See Table 3.1) on page 54, where the volume of FCC crystal structure is given by equation 3.6 by
$$\displaystyle{V}_{{c}}={16}{R}^{{3}}\sqrt{{2}}\ \ \ {\left({1}\right)}$$
Now we can plug the vlaue for R in equation (1) to get $$\displaystyle{V}_{{c}}$$ where R will be in (m)
$$\displaystyle{V}_{{c}}={16}{R}^{{3}}\sqrt{{2}}$$
$$\displaystyle={16}{\left({0.175}\times{10}^{{-{9}}}{m}\right)}^{{3}}\sqrt{{2}}$$
$$\displaystyle={1.21}\times{10}^{{-{{28}}}}{m}^{{3}}$$
Result
$$\displaystyle{V}_{{c}}={1.21}\times{10}^{{-{{28}}}}{m}^{{3}}$$