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Wave Nature of Matter



Wave Nature Of Matter (de Broglie Hypothesis $ \lambda = h/p $)

The success of Einstein's photon theory in explaining the photoelectric effect established the particle nature of light. This, along with the well-known wave phenomena of light (interference, diffraction), led to the concept of wave-particle duality for light. In 1924, Louis de Broglie, a French physicist, proposed a radical idea: if light exhibits both wave and particle properties, then perhaps matter (like electrons, protons, atoms, etc.), which we traditionally consider particles, also exhibits wave-like properties. This is known as the wave nature of matter or de Broglie hypothesis.


De Broglie Hypothesis

De Broglie's hypothesis states that:

"A moving material particle, whatever its nature, has an associated wave, called the de Broglie wave or matter wave."

He proposed that the wavelength ($\lambda$) of this matter wave is related to the momentum ($p$) of the particle by the same relation that holds for photons:

$ \lambda = \frac{h}{p} $

Where $h$ is Planck's constant.

For a particle of mass $m$ moving with velocity $v$, its momentum is $p = mv$. Thus, the de Broglie wavelength is:

$ \lambda = \frac{h}{mv} $

This equation is the mathematical statement of the de Broglie hypothesis. It connects the wave property (wavelength $\lambda$) with the particle property (momentum $p = mv$).

De Broglie's idea was purely theoretical at the time. There was no direct experimental evidence for the wave nature of electrons or other particles.


De Broglie Wavelength for Different Particles

The de Broglie wavelength depends inversely on the momentum of the particle.


De Broglie Wavelength of an Electron Accelerated by a Potential Difference

A common way to give kinetic energy to an electron is by accelerating it through a potential difference. If an electron (charge $e$, mass $m$) is accelerated from rest through a potential difference $V$, the work done on it is $eV$. This work is converted into kinetic energy $K$.

$ K = \frac{1}{2} mv^2 = eV $

We can express the momentum $p = mv$ in terms of kinetic energy:

$ p^2 = (mv)^2 = m^2 v^2 = 2m \left(\frac{1}{2}mv^2\right) = 2mK $

$ p = \sqrt{2mK} $

For an electron accelerated by potential difference $V$, $K = eV$.

$ p = \sqrt{2meV} $

The de Broglie wavelength of the accelerated electron is:

$ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}} $

Substituting the values of $h, m, e$: $h = 6.626 \times 10^{-34} \, J \cdot s$, $m_e = 9.109 \times 10^{-31} \, kg$, $e = 1.602 \times 10^{-19} \, C$.

$ \sqrt{2me} = \sqrt{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19}} = \sqrt{29.18 \times 10^{-50}} \approx 5.4 \times 10^{-25} $

$ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{V} \times 5.4 \times 10^{-25}} \, m $

$ \lambda \approx \frac{1.22 \times 10^{-9}}{\sqrt{V}} \, m = \frac{1.22}{\sqrt{V}} \, nm $

This formula gives the de Broglie wavelength of an electron accelerated through a potential difference $V$. For example, an electron accelerated through 100 V has a wavelength of about 0.122 nm, which is comparable to the spacing between atoms in crystals. This is why diffraction effects for electrons can be observed using crystal lattices, as demonstrated by the Davisson-Germer experiment.


Significance of De Broglie Hypothesis

De Broglie's hypothesis of matter waves was a bold and revolutionary idea that extended the wave-particle duality from light to matter. It provided a theoretical basis for understanding the behaviour of particles at the atomic and subatomic levels. The experimental verification of this hypothesis (Davisson-Germer experiment for electrons) confirmed the wave nature of matter and was a cornerstone in the development of quantum mechanics. Quantum mechanics treats particles not just as localised points, but as entities described by wave functions whose properties (like probability of finding the particle) are related to the wave's characteristics.


Example 1. Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 500 V. ($h = 6.63 \times 10^{-34} \, J \cdot s$, $m_e = 9.11 \times 10^{-31} \, kg$, $e = 1.60 \times 10^{-19} \, C$).

Answer:

Given:

Potential difference, $V = 500 \, V$

Planck's constant, $h = 6.63 \times 10^{-34} \, J \cdot s$

Mass of electron, $m_e = 9.11 \times 10^{-31} \, kg$

Charge of electron, $e = 1.60 \times 10^{-19} \, C$

The de Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.

Substitute the values:

$ \lambda = \frac{6.63 \times 10^{-34} \, J \cdot s}{\sqrt{2 \times (9.11 \times 10^{-31} \, kg) \times (1.60 \times 10^{-19} \, C) \times (500 \, V)}} $

$ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 1.60 \times 500 \times 10^{-31} \times 10^{-19}}} \, m $

$ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{14576 \times 10^{-50}}} \, m = \frac{6.63 \times 10^{-34}}{\sqrt{1.4576 \times 10^{-46}}} \, m $

$ \lambda = \frac{6.63 \times 10^{-34}}{1.207 \times 10^{-23}} \, m $

$ \lambda \approx 5.49 \times 10^{-11} \, m $

We can also use the simplified formula $\lambda \approx \frac{1.22}{\sqrt{V}} \, nm$:

$ \lambda \approx \frac{1.22}{\sqrt{500}} \, nm = \frac{1.22}{22.36} \, nm \approx 0.0545 \, nm $

$ \lambda \approx 0.0545 \times 10^{-9} \, m = 5.45 \times 10^{-11} \, m $

The results from the detailed calculation and the simplified formula are close, the slight difference is due to the values of constants used in the simplified formula. The de Broglie wavelength of the electron is approximately $5.49 \times 10^{-11}$ meters or 0.0549 nm. This wavelength is in the range of atomic sizes and crystal lattice spacings, making diffraction observable.