A typical atom has a diameter of about\({\bf{1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\;{\bf{m}}\).

(a) What is this in inches?

(b) Approximately how many atoms are along a 1.0 cm line, assuming they just touch?

The unit conversion is a process in which the unit of a particular quantity is changed from one standard to another standard by multiplying or dividing by a fixed factor.

Given data:

The diameter of a typical atom is\(1.0 \times {10^{ - 10}}\;{\rm{m}}\).

Now,

\(\begin{aligned}{c}{\rm{1}}\;{\rm{in}} = 2.54\;{\rm{cm}}\\ = \,2.54\;{\rm{cm}}\, \times \,\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}\\ = 0.0254\,{\rm{m}}\end{aligned}\).

(a) Now, changing the unit from m to in, you get

\(\begin{aligned}{c}1.0 \times {10^{ - 10}}{\rm{m}} = 1.0 \times {10^{ - 10}}\;{\rm{m}}\, \times \frac{{\,1\;{\rm{in}}}}{{0.0254\;{\rm{m}}}}\\ = 3.9 \times {10^{ - 9}}\;{\rm{in}}\end{aligned}\).

Therefore, the diameter of a typical atom is about \(3.9 \times {10^{ - 9}}\;{\rm{in}}\).

(b)Let there be N atoms along a 1.0-cm line.

Now, you know that \(1.0\;{\rm{cm}} = 1.0 \times {10^{ - 2}}\;{\rm{m}}\).

Assuming they just touch together, you get

\(\begin{aligned}{c}N \times \left( {1.0 \times {{10}^{ - 10}}\;{\rm{m}}} \right) = 1.0 \times {10^{ - 2}}\;{\rm{m}}\\N = \frac{{1.0 \times {{10}^{ - 2}}\;{\rm{m}}}}{{1.0 \times {{10}^{ - 10}}\;{\rm{m}}}}\\N = {10^8}\end{aligned}\)

Therefore, there are 10⁸atoms along a 1.0 cm line.