The binding energy of atoms is the minimum energy required to remove an electron of an atom from its orbital. Using dimensional analysis, we can derive the following equation. $$E_B\propto\frac{m(ke^2)^2}{\hbar^2}$$ When I was considering parameter combinations (includuing the speed of light, $c$). I came across a dimensional quantity by combining the parameters $e$, $c$, $k$ and $\hbar$. $$\alpha = \frac{ke^2}{\hbar c}$$ Naturally, I turned to google and learnt about this quantity called the fine structure constant given as $\alpha$.The fine structure constant is used to define relativistic corrections to the binding energy of the atom. Now the relativistic equation of the binding energy of an atom is given as the following.$$E_B\propto \frac{m(ke^2)^2}{\hbar^2}f(\alpha)$$ Here the function, $f(\alpha)=1+\frac{5}{4}\alpha^2$.

Therefore my obvious queries are -

  1. Why is $f(\alpha)=1+\frac{5}{4}\alpha^2$?
  2. Are there any other relativistic corrections to the equation?

1 Answer 1


The full answer is a long story; I will give the highlights.

First, if the electrostatic force between point charges is $$ f = k \frac{q_1 q_2}{r^2} $$ (Coulomb's law) then the constant $k$ is equal to $1/(4 \pi \epsilon_0)$ in the SI units for electromagnetism. Since these are often adopted, we write the fine structure constant as $$ \alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c}. $$

The binding energy of hydrogen is obtained by solving the quantum mechanical description of two point particles with opposite charges. A first approach adopts non-relativistic quantum theory, and in this case the answer is $$ E = \frac{1}{2} \mu \alpha^2 c^2 $$ where $\mu$ (called the reduced mass) is $$ \mu = \frac{m_e m_p}{m_e + m_p} $$ in which $m_e$ and $m_p$ are the masses of electron and proton respectively. To answer the question 'why this answer?' it would require one to learn quantum mechanics. Basically it is a combination of kinetic energy and electrostatic potential energy, which comes about as a kind of equilibrium where the if the electron were on average any closer to the proton (thus lowering the potential energy) then the kinetic energy would have to increase. Another way to say the same thing is to assert that the kinetic energy influences the wavelength of the wavefunction of the electron, such that a small wavelength implies a higher momentum and hence a higher kinetic energy.

Anyway that is the non-relativistic result.

Relativistic corrections come in three steps. First one can modify the non-relativistic theory by adding corrections to the formula for kinetic energy and things like that. But a better way, the second step, is to replace Schrodinger's equation with Dirac's equation for the system. Dirac's equation takes relativity into account right from the start. It gives the answer $$ E = \frac{1}{2} \mu \alpha^2 c^2 f(\alpha,\, j) $$ where $f(\alpha, ,j)$ depends on $\alpha$ and the total angular momentum, indicated by a quantum number called $j$. This function can be written as a power series expansion in powers of $\alpha^2$.

Finally, the full modern treatment is much more complicated and has to adopt the ideas and methods of quantum field theory, in this case quantum electrodynamics or QED. The calculation involves a sequence of integrals which account for all possible ways the electron and proton can exchange energy and momentum. One interesting feature is that according to QED, some of the other energy levels of hydrogen (not the ground state) sharing the same $j$ now appear in pairs split by a tiny gap called the Lamb shift. This effect also influences the ground state energy and therefore the binding energy.

The binding energy of hydrogen is now one of the most accurately measured quantities in all of science, and its calculation is one of the most stringent tests of QED. The agreement between theory and experiment amounts to a test of the whole structure of the theory of QED; one no longer speaks of relativistic corrections, but rather a complete structure of ideas.


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