How fast do electrons move around the nucleus?
The ratio of the speed of an electron traveling in the first Bohr orbit to the speed of light is given by the handy equation

$$\mathrm{V_{rel}=\frac{[Z]}{[137]}}$$

where Z is the atomic number of the element under consideration and 137 is the speed of light in atomic units, also known as the fine structure constant.  Consequently a 1s electron in the hydrogen atom will travel at approximately 0.7% the speed of light. In silver (Z=47) the 1s electron will travel around 34% the speed of light, while the 1s electron in gold (Z=79) will travel at about 58% the speed of light.  

Once we get up to around silver, the electrons are traveling at relativistic speeds and this can dramatically impact properties of the atom. For example, the relativistic mass of an electron is given by

$$\mathrm{m_{rel}=\frac{m_{e}}{\sqrt{1-(V_{rel}/c)^2}}}$$

where $\ce{m_{e},~ V_{rel}~ and~ c}$ are the electron rest mass, the velocity of the electron, and the speed of light respectively.  The following figure provides a graphical representation of how the electron mass increases as the electron velocity increases.



The following equation relates the ratio of the relativistic radius of the first Bohr orbit $\ce{R_{rel}}$ to the normal radius $\ce{R_{o}}$, to the relativistic velocity of the electron

$$\mathrm{\frac{[R_{rel}]}{[R_{o}]}=\sqrt{1-(V_{rel}/c)^2}}$$

As the relativistic velocity of the electron increases, the orbital radius contracts (the above ratio becomes smaller).  For silver, the first Bohr radius contracts ~6%, while for gold the contraction is ~ 18%.

Look at these earlier Chem SE answers to see the interesting physical effects that atoms can exhibit when their electrons travel at relativistic speeds.


francium
gold
mercury