I know that for atoms with big values of Z, relativistic effects have to be applied. Why does electron velocity increase with the increase of atomic number?


You can easily understand that based on the particle in the box model. Large Z corresponds to large electrostatic field around the nucleus which act on the electrons. This leads to a smaller "box", ie. the core electrons of heavy elements are moving on orbitals with radius orbitals than electrons of light elements. Since the energy of the electron is (if we use the simple 1D version) $$E = \frac{n^2h^2}{8mL^2}$$ where $L$ corresponds to the box size and the energy ($E$) is purely kinetic energy. You can see that smaller box corresponds to larger kinetic energies.


Why does electron velocity increase with the increase of atomic number?

Newton's second law tells us that for an electron moving in a stable circular orbit around a nucleus the centripetal force pushing out must equal the electrostatic attraction pulling in, or

$${\frac{m_e~ v^2}{r}} = {\frac{Z~k_e ~e^2}{r^2}}~~(1)$$

where $m_e$ is the electron's mass, e is the electron charge, $k_e$ is Coulomb's constant and $Z$ is the atom's atomic number. Rearranging equation (1) yields

$$v = \sqrt{ Z~k_e~ e^2 \over m_\mathrm{e} r}~~(2) $$

From equation (2) we can see that the electron velocity increases as the atomic number, $Z$, increases.

Edit: Response to OP's comment

I am expecting some quantum mechanical explanation

As the electron velocity given by equation (2) becomes comparable to the speed of light, the mass of the electron increases due to relativistic effects.


As the electron mass increases, the s (and somewhat for p) orbital radius contracts. So relativistic effects affect the electron mass and the orbital radius. In terms of electron velocity, these 2 relativistic effects offset one another and the electron velocity is not affected; equation (2) is still valid.

  • $\begingroup$ I am expecting some quantum mechanical explanation. $\endgroup$
    – EJC
    Nov 6 '14 at 17:03
  • $\begingroup$ The quantum mechanical response is similar. The potential energy of the electron includes the electrostatic interaction between the nuclear charge Z and the electron. Even ignoring the "stable circular orbit" idea, there is a balance between the angular momentum and the electrostatic attraction and the remaining equations are similar. $\endgroup$ Nov 6 '14 at 20:45

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