Question

An atom of tin ({\rm Sn}) has a diameter of about 2.8\times10^{-8}\;{\rm cm}.
How many {\rm Sn} atoms would have to be placed side by side to span a distance of 7.0\mu m ?

Answer 1

An atom of tin $(\{\backslash \mathrm{rm} \ \mathrm{Sn}\})$ has a diameter of about 2.8\times10^{-8}\;(\backslash \mathrm{rm} \ \mathrm{cm})$. How many$\{\backslash \mathrm{rm} \ \mathrm{Sn}\}$ atoms would have to be placed side by side to span a distance of 7.0\mu m?

**Solution:**

One meter include $100\ \mathrm{cm}$, so that in distance of $700\ \mathrm{cm}$ be able to place side by side: $700 / 2.8 \cdot 10^{-8} = 2.5 \cdot 10^{10}$ in number of atoms of tin.

**Answer:** $2.5 \cdot 10^{10}$ atoms of tin.