What is the amplitude of the electron wave when it is orbiting nucleus?

Question

We all know that electrons have wave properties and we can calculate the de Broglie wavelength. But what about the amplitude of the wave?

Answer 1

We should start by noting that the de Broglie wavelength is only well defined for a free particle. We should also note that the electron in an atom doesn't orbit like a planet around a star. If they did they would quickly spiral into the nucleus and crash!

An electron is described by its wave function, $\psi$. For a free particle the wave function turns out to be:

$$ \psi(t,x) = e^{i(kx-\omega t)} \tag{1} $$

and this is just the equation of a plane wave so it has a well defined wavelength. The de Broglie wavelength $\lambda$ is related to the $k$ in the equation by:

$$ k = \frac{2\pi}{\lambda} $$

For an electron in a hydrogen atom the wavefunction in the ground state is:

$$ \psi(r) = \frac{2}{a^{3/2}} e^{-r/a} \tag{2} $$

where $a$ is the Bohr radius. This doesn't look like a plane wave and has no de Broglie wavelength.

But we can still answer your question because the amplitude of the wave function is related to the probability of finding the particle. Specifically the probability of finding the particle in an infinitesimal volume $dV$ is given by:

$$ P = |\psi|^2dV $$

And since the probability of finding the particle somewhere must be unity we know that:

$$ \int |\psi|^2dV = 1 $$

This condition is called normalisation, and it's how the prefactor of $2/a^{3/2}$ was calculated in equation (2).