Section 54.1 Matter Waves
Louis De Broglie in France had a breakthrough in thinking when he asked: if a photon can be have both particle and wave properties, why can't an electron also have particle and wave properties? He thought that, if electrons were waves, one could understand the stability of electronic orbits, called Bohr orbits (to be addressed in the next chapter), similar to the stability of standing waves on a string. For instance, the size of a Bohr orbit could correspond to an integral multiple of the wavelength of the electron wave as illustrated in Figure 54.1.1.
demonstrations.wolfram.comThus, for a Bohr orbit of radius \(r\text{,}\) there will be a natural quantization:
where \(\lambda\) would be the wavelength of the electron wave and \(n\) some integer. But, what should be the wavelength of an electron? He used the analogy with photon and proposed a parallel concept for electrons as shown in Table 54.1.2.
| Property | Photon | Electron |
|---|---|---|
| Momentum | \(p = \dfrac{h}{\lambda}\) | \(p = \dfrac{h}{\lambda} \leftrightarrow \lambda = \dfrac{h}{p}.\) |
| Energy | \(E = p c = h f\) | \(E = \sqrt{p^2c^2 + m^2c^4}\) |
| \(\ \ = \dfrac{hc}{\lambda} \) | \(\ \ = \sqrt{h^2c^2/\lambda^2 + m^2c^4}.\) | |
With this assumption de Broglie could account for the discrete spectrum of Hydrogen atom quite naturally. With the success of thinking electrons as waves de Broglie stated that the wave picture should apply to all material particles.
The question that we can now ask is: Why do we not see wave-like properties of ordinary objects? To that question, de Broglie's answer is that their wavelengths are too small compared to the dimensions of the object. Therefore, even though they are waves, we cannot see their wave aspects so easily. In the following we will work out some examples of de Broglie wavelengths.
Checkpoint 54.1.3. Wavelength of an Electron.
What is the de Broglie wavelength of an electron moving at \(1.0\times 10^{5}\:\textrm{m/s}\text{?}\)
Use momentum.
\(7.3\:\textrm{nm}\text{.}\)
First we need to decide if we will use the relativistic formula or the non-relativistic formula for calculating the momentum. Here we have \(v/c = 0.33\times 10^{-3}\text{,}\) which is much very small compared to \(1\text{.}\) Therefore, we will use the non-relativistic formula.
Therefore, the wavelength of the electron will be
Checkpoint 54.1.4. Wavelength of a Ball.
What is the de Broglie wavelength of a ball of mass \(0.350\:\textrm{kg}\) moving at \(45.0\:\textrm{m/s}\text{?}\)
Use momentum.
\(4.2\times 10^{-35}\:\textrm{m}\text{.}\)
This is a non-relativistic situation.
Therefore, the wavelength of the electron will be
Clearly, this is too small a wavelength for the wave picture to be of much use.
Checkpoint 54.1.5. Wavelength of a Relativistic Electron.
What is the de Broglie wavelength of an electron which has a kinetic energy of \(100\:\textrm{keV}\text{?}\)
Use relativistic energy formula.
\(5.9\times 10^{-13}\:\textrm{m}\text{.}\)
Rest energy of an electron is \(511\text{ keV}\text{.}\) Since the kinetic energy is not too small compared to the rest energy, the particle should be treated relativistically.
Therefore, \(pc\) will be
Therefore, the wavelength will be
Now, if we use the value of \(hc\) in units \(\textrm{eV.m}\) and the energies in \(\textrm{eV}\) we will not have to change too many units.
and
Therefore,