The radius of an atom of tungsten \((\mathrm{W})\) is about \(2.10 \AA\) (a) Express this distance in nanometers (nm) and in picometers (pm). (b) How many tungsten atoms would have to be lined up to create a wire of $2.0 \mathrm{~mm}$ (c) If the atom is assumed to be a sphere, what is the volume in \(\mathrm{m}^{3}\) of a single \(\mathrm{W}\) atom?

To convert the given radius in angstroms to nanometers and picometers, we will use the conversion factors: 1 angstrom (Å) = 0.1 nanometers (nm) 1 angstrom (Å) = 100 picometers (pm) Given radius: 2.10 Å Convert to nanometers: \(2.10 \mathrm{Å} \times \frac{0.1 \mathrm{nm}}{1 \mathrm{Å}} = 0.210 \mathrm{nm}\) Convert to picometers: \(2.10 \mathrm{Å} \times \frac{100 \mathrm{pm}}{1 \mathrm{Å}} = 210 \mathrm{pm}\) So, in nanometers the radius is \(0.210 \mathrm{nm}\) and in picometers the radius is \(210 \mathrm{pm}\).

To find the number of tungsten atoms required to create a 2 mm long wire, we'll first convert the length of the wire to the same units as the radius (in picometers), and then divide the length of the wire by the diameter of a single W atom (which is twice the radius). Convert wire length to picometers: \(2.0 \mathrm{mm} \times \frac{10^{9} \mathrm{pm}}{1 \mathrm{mm}} = 2 \times 10^{9} \mathrm{pm}\) Diameter of a single W atom = 2 × radius = \(2 \times 210 \mathrm{pm} = 420 \mathrm{pm}\) Number of W atoms in a 2 mm wire: \(\mathrm{Number\ of\ W\ atoms}=\frac{2\times 10^{9} \mathrm{pm}}{420 \mathrm{pm}}=4.76 \times 10^6\) So, approximately \(4.76 \times 10^6\) W atoms are needed to create a 2 mm long wire.

To calculate the volume of a single tungsten atom, we assume it to be a sphere and use the formula: \(V = \frac{4}{3} \pi r^{3}\) Here, V represents the volume, r represents the radius, and \(\pi\) is a constant with an approximate value of 3.14159. The radius must be in meters for the volume to be in \(\mathrm{m}^{3}\). Convert radius to meters: \(r = 2.10 \mathrm{Å} \times \frac{0.1 \mathrm{nm}}{1 \mathrm{Å}} \times \frac{1 \times 10^{-9} \mathrm{m}}{1 \mathrm{nm}} = 2.10 \times 10^{-10} \mathrm{m}\) Volume of a single W atom: \(V = \frac{4}{3} \pi (2.10 \times 10^{-10} \mathrm{m})^{3}\) \(V = 4.086 \times 10^{-29} \mathrm{m}^{3}\) Thus, the volume of a single W atom is approximately \(4.086 \times 10^{-29} \mathrm{m}^{3}\).