11.8 Wave Nature of Matter

The dual (wave-particle) nature of light (electromagnetic radiation, in general) comes out clearly from what we have learnt in this and the preceding chapters. The wave nature of light shows up in the phenomena of interference, diffraction and polarisation. On the other hand, in photoelectric effect and Compton effect which involve energy and momentum transfer, radiation behaves as if it is made up of a bunch of particles – the photons. Whether a particle or wave description is best suited for understanding an experiment depends on the nature of the experiment. For example, in the familiar phenomenon of seeing an object by our eye, both descriptions are important. The gathering and focussing mechanism of light by the eye-lens is well described in the wave picture. But its absorption by the rods and cones (of the retina) requires the photon picture of light.

A natural question arises: If radiation has a dual (wave-particle) nature, might not the particles of nature (the electrons, protons, etc.) also exhibit wave-like character? In 1924, the French physicist Louis Victor de Broglie (pronounced as de Broy) (1892-1987) put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions. He reasoned that nature was symmetrical and that the two basic physical entities – matter and energy, must have symmetrical character. If radiation shows dual aspects, so should matter. De Broglie proposed that the wavelength \(\lambda\) associated with a particle of momentum \(p\) is given as
\(
\lambda=\frac{h}{p}=\frac{h}{m v} \dots(11.5)
\)
where \(m\) is the mass of the particle and \(v\) its speed. Equation (11.5) is known as the de Broglie relation and the wavelength \(\lambda\) of the matter wave is called de Broglie wavelength. The dual aspect of the matter is evident in the de Broglie relation. On the left-hand side of Eq. (11.5), \(\lambda\) is the attribute of a wave while on the right-hand side, the momentum \(p\) is a typical attribute of a particle. Planck’s constant \(h\) relates the two attributes.

Equation (11.5) for a material particle is basically a hypothesis whose validity can be tested only by experiment. However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen,
\(
p=h \nu / c \dots(11.6)
\)
Therefore, \(\frac{h}{p}=\frac{c}{\nu}=\lambda \dots(11.7)\)
That is, the de Broglie wavelength of a photon given by Eq. (11.5) equals the wavelength of electromagnetic radiation of which the photon is a quantum of energy and momentum.

Clearly, from Eq. (11.5), \(\lambda\) is smaller for a heavier particle (large \(m\) ) or more energetic particle (large \(v\) ). For example, the de Broglie wavelength of a ball of mass 0.12 kg moving with a speed of \(20 \mathrm{~m} \mathrm{~s}^{-1}\) is easily calculated:
\(
p=m v=0.12 \mathrm{~kg} \times 20 \mathrm{~m} \mathrm{~s}^{-1}=2.40 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
\)
\(
\lambda=\frac{h}{p}=\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{2.40 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}}=2.76 \times 10^{-34} \mathrm{~m}
\)
This wavelength is so small that it is beyond any measurement. This is the reason why macroscopic objects in our daily life do not show wavelike properties. On the other hand, in the sub-atomic domain, the wave character of particles is significant and measurable.

Example 11.3: What is the de Broglie wavelength associated with (a) an electron moving with a speed of \(5.4 \times 10^6 \mathrm{~m} / \mathrm{s}\), and (b) a ball of mass 150 g travelling at \(30.0 \mathrm{~m} / \mathrm{s}\)?

Solution: (a) For the electron:
Mass \(m=9.11 \times 10^{-31} \mathrm{~kg}\), speed \(v=5.4 \times 10^6 \mathrm{~m} / \mathrm{s}\). Then, momentum
\(
\begin{aligned}
& p=m v=9.11 \times 10^{-31}(\mathrm{~kg}) \times 5.4 \times 10^6(\mathrm{~m} / \mathrm{s}) \\
& p=4.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
de Broglie wavelength, \(\lambda=h / p\)
\(
\begin{aligned}
& =\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{4.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}} \\
\lambda & =0.135 \mathrm{~nm}
\end{aligned}
\)
(b) For the ball:
Mass \(m^{\prime}=0.150 \mathrm{~kg}\), speed \(v^{\prime}=30.0 \mathrm{~m} / \mathrm{s}\).
Then momentum \(p^{\prime}=m^{\prime} v^{\prime}=0.150(\mathrm{~kg}) \times 30.0(\mathrm{~m} / \mathrm{s})\)
\(
p^{\prime}=4.50 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\)
de Broglie wavelength \(\lambda^{\prime}=h / p^{\prime}\).
\(
\begin{aligned}
& =\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{4.50 \times \mathrm{kg} \mathrm{~m} / \mathrm{s}} \\
& \lambda^{\prime}=1.47 \times 10^{-34} \mathrm{~m}
\end{aligned}
\)
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about \(10^{-19}\) times the size of the proton, quite beyond experimental measurement.