De Broglie Hypothesis

The hypothesis

In 1924, Prince Louis de Broglie proposed, on the basis of the symmetry existing in nature, that if radiation has a dual nature — sometimes wave nature dominating, sometimes particle nature — then matter may also possess a dual nature.

• Normally we treat matter as composed of particles, but there should be situations in which matter shows wave-like properties. The diffraction of electrons by nickel crystals, described in the next section, is exactly such a situation.
• De Broglie used the wave properties frequency $\nu$ and wavelength $\lambda$ and proposed a relation connecting them with the particle properties energy $E$ and momentum $p$.

Starting point: the momentum of a photon

The momentum carried by a photon of energy $E$ is

$$p = \dfrac{E}{c}$$

which is valid for a massless particle travelling with the speed of light $c$, according to Einstein's special theory of relativity. Using Einstein's relation $E = h\nu$ and $\lambda\nu = c$,

$$p = \dfrac{E}{c} = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}$$

De Broglie's proposal

De Broglie proposed that a moving material particle of total energy $E$ and momentum $p$ has associated with it a wave, analogous to a photon, and that the wave and particle properties of matter can be described by a relation similar to the one above. So for a particle of mass $m$ moving with velocity $v$,

$$\nu = \dfrac{E}{h} \qquad \text{and} \qquad \lambda = \dfrac{h}{p} = \dfrac{h}{mv}$$

• He called these waves associated with material particles matter waves, and $\lambda$ is the de Broglie wavelength. The equation is the de Broglie relation.
• Greater the momentum, shorter the wavelength.
• An electron at rest has $v = 0$, hence $p = 0$, hence $\lambda = h/p$ is infinite — that is, undefined. Only a moving electron has an associated de Broglie wavelength.

Wavelength in terms of kinetic energy

For a particle of mass $m$ moving with velocity $v$, the kinetic energy is $E_K = \dfrac{1}{2}mv^2$, so $v = \sqrt{2E_K/m}$. Substituting,

$$\lambda = \dfrac{h}{mv} = \dfrac{h}{m\sqrt{2E_K/m}} = \dfrac{h}{\sqrt{2mE_K}}$$

Equivalently, since $p = mv$, we have $p^2 = m^2v^2$ and

$$\left(\dfrac{1}{2}mv^2\right)m = \dfrac{1}{2}p^2 \qquad \therefore \ E_K = \dfrac{p^2}{2m} \qquad \therefore \ p = \sqrt{2mE_K}$$

$$\therefore \ \lambda = \dfrac{h}{\sqrt{2mE_K}}$$

Derivation for a charged particle accelerated through V volt

This is the derivation the board sets. Work through it in these steps.

• Consider a particle of mass $m$ and charge $q$ accelerated from rest through a potential difference of $V$ volt.
• By the de Broglie hypothesis, $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$, with $p = mv$.
• Since $E_K = \dfrac{p^2}{2m}$, we get $p = \sqrt{2mE_K}$ and hence $\lambda = \dfrac{h}{\sqrt{2mE_K}}$.
• The work done on the particle by the accelerating field is stored as its kinetic energy, so $E_K = qV$. Therefore

$$\lambda = \dfrac{h}{\sqrt{2mE_K}} = \dfrac{h}{\sqrt{2mqV}}$$

• For an electron, substituting $q = 1.6\times10^{-19}$ C, $m = 9.1\times10^{-31}$ kg and $h = 6.63\times10^{-34}$ J s,

$$\lambda = \dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times V(\text{in volt})}} = \dfrac{1.228\times10^{-9}}{\sqrt{V(\text{in volt})}}\ \text{m}$$

$$\therefore \ \lambda = \dfrac{1.228}{\sqrt{V(\text{in volt})}}\ \text{nm} = \dfrac{12.28}{\sqrt{V(\text{in volt})}}\ \text{angstrom}$$

• This relation holds for any charged particle — electron, proton or charged ion — with $m$ the mass of that particle. It fails when $V$ is very large, say in kilovolt, because then the speed approaches the speed of light and relativistic corrections are needed.

Why we never see matter waves of everyday objects

• A student of mass $45$ kg running at $8\ \text{km h}^{-1}$ has momentum $100\ \text{kg m s}^{-1}$, so $\lambda = 6.63\times10^{-36}$ m.
• A car of mass $1200$ kg moving at $60\ \text{km h}^{-1}$ has momentum $20000\ \text{kg m s}^{-1}$, so $\lambda = 3.32\times10^{-38}$ m.
• These wavelengths are utterly negligible compared with the size of the moving object and with the width of the path — the $2$ m footpath and the $20$ m road. They also cannot be measured. So the wave nature of macroscopic bodies plays no role in everyday life.
• By contrast, an electron of kinetic energy $100$ eV has speed about $5.93\times10^{7}\ \text{m s}^{-1}$, momentum $5.40\times10^{-24}\ \text{kg m s}^{-1}$ and $\lambda = 1.23\times10^{-10}$ m, that is $1.228$ angstrom. That is comparable with the diameter of a $2$ angstrom hole it passes through, the same as the size of a helium atom, and more than double the size of a hydrogen atom — so its wave nature is readily observable.
• The wavelength associated with an electron of a few eV is of the order of a few angstrom, so to observe the wave nature of electrons the slit width or the diffracting object must be of the same order of magnitude.
• Sub-atomic particles like electrons, protons and neutrons, and atomic particles too, all exhibit wave properties. The wave property of the electron was confirmed by Davisson and Germer in 1927 and by G. P. Thomson in 1928.

Standard results collected

$$\lambda = \dfrac{h}{p} = \dfrac{h}{mv} \qquad \nu = \dfrac{E}{h} \qquad \lambda = \dfrac{h}{\sqrt{2mE_K}} \qquad \lambda = \dfrac{h}{\sqrt{2mqV}}$$

$$\lambda = \dfrac{1.228}{\sqrt{V(\text{in volt})}}\ \text{nm} \qquad E_K = \dfrac{p^2}{2m} \qquad v = \sqrt{\dfrac{2E_K}{m}}$$

• Equal speeds: $\lambda \propto 1/m$, so for an electron and a proton $\lambda_e/\lambda_p = 1836$, electron longer.
• Equal kinetic energies: $\lambda \propto 1/\sqrt{m}$, so $\lambda_e/\lambda_p = \sqrt{1836} = 42.85$, electron longer. Among an electron, a proton, an alpha particle and a hydrogen atom of the same kinetic energy, the electron has the longest wavelength.
• Equal momenta: wavelengths are equal.
• Equal wavelengths: momenta are equal, and $E_e/E_p = m_p/m_e = 1836$.
• Two particles of equal wavelength with one moving four times as fast as the other must satisfy $m_1v_1 = m_2v_2$; if the slower one is an alpha particle of mass $4$ units, the faster one has mass $1$ unit, so it is a proton or a neutron.