5.5: Wave-Particle Duality of Matter
- Page ID
- 465528
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)So far we have seen that to fully understand some experimental results with light we cannot simply treat them as waves but sometimes we have to consider that they also have properties of particles (photons). It turns out that for particles like the electrons, there are some experimental results whose results can only be understood if we treat those particles as waves.
Light Waves
Many of the properties of light (e.g., surface of water or acoustics) including reflection, refraction, diffraction, and interference can be explained, both qualitatively and quantitatively, in terms of light viewed as a wave.
![ezgif-4-085bec536432.gif](https://chem.libretexts.org/@api/deki/files/241029/ezgif-4-085bec536432.gif?revision=1)
The amplitudes of waves add. Constructive interference is obtained when identical waves are in phase and destructive interference occurs when identical waves are exactly out of phase, or shifted by half a wavelength.
![Two_sources_interference.gif](https://chem.libretexts.org/@api/deki/files/85400/Two_sources_interference.gif?revision=1)
Quantum Matter Waves
Continuing with our analysis of experiments that lead to the new quantum theory, we now look at the phenomenon of electron diffraction. It is well-known that light has the ability to diffract around objects in its path, leading to an interference pattern that is particular to the object. This is, in fact, how holography works (the interference pattern is created by allowing the diffracted light to interfere with the original beam so that the hologram can be viewed by shining the original beam on the image). A simple illustration of diffraction is the Young double slit experiment pictured below:
![CNX_UPhysics_36_01_YoungSetUp.jpg](https://chem.libretexts.org/@api/deki/files/122916/CNX_UPhysics_36_01_YoungSetUp.jpg?revision=1)
Amazingly, if electrons are used instead of light in the double-slit experiment, and a fluorescent screen is used, one finds the same kind of interference pattern! This is shown in the electron double-slit diffraction pattern below:
![512px-Double-slit.svg.png](https://chem.libretexts.org/@api/deki/files/122917/512px-Double-slit.svg.png?revision=1)
Obviously, classical mechanics is not able to predict such a result. If the electrons are treated as classical particles, one would predict an intensity pattern corresponding to particles that can pass through one slit or the other, landing on the screen directly opposite the slit (i.e., no intensity maximum at the center of the screen):
Through such experiments, the idea that electrons can behave as waves, creating interference patterns normally associated with light, is now well-established. The fact that particles can behave as waves but also as particles, depending on which experiment you perform on them, is known as the particle-wave duality.
Deriving the De Broglie Wavelength
De Broglie derived his equation using well established theories through the following series of substitutions:
1. De Broglie first used Einstein's famous equation relating matter and energy:
\[ E = mc^2 \label{0}\]
with
- \(E\) = energy,
- \(m\) = mass,
- \(c\) = speed of light
2. Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation:
\[ E= h \nu \label{1}\]
with
- \(E\) = energy,
- \(h\) = Plank's constant (6.62607 x 10-34 J s),
- \(\nu\)= frequency
3. Since de Broglie believed particles and wave have the same traits, he hypothesized that the two energies would be equal:
\[ mc^2 = h\nu \label{2}\]
4. Because real particles do not travel at the speed of light, De Broglie submitted velocity (\(v\)) for the speed of light (\(c\)).
\[ mv^2 = h\nu \label{3}\]
5. Since the speed of a wave relates its frequency to its wavelength, de Broglie substituted \( v/\lambda\) for \(\nu\) and arrived at the final expression that relates wavelength and particle with speed.
\[ mv^2 = \dfrac{hv}{\lambda} \label{4}\]
Hence:
\[ \lambda = \dfrac{hv}{mv^2} = \dfrac{h}{mv}= \dfrac{h}{p} \label{5} \]
At this stage, a majority of Wave-Particle Duality problems are simply "plug and chug" using Equation \ref{5} with some variation of canceling out units
Calculate the wavelength of a baseball, which has a mass of 149 g and a speed of 100 mi/h.
Given: mass and speed of object
Asked for: wavelength
Strategy:
- Convert the speed of the baseball to the appropriate SI units: meters per second.
- Substitute values into Equation \(\ref{5}\) and solve for the wavelength.
Solution:
The wavelength of a particle is given by \(λ = h/mv\). We know that m = 0.149 kg, so all we need to find is the speed of the baseball:
\( v=\left ( \dfrac{100\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right ) \)
B Recall that the joule is a derived unit, whose units are (kg•m2)/s2. Thus the wavelength of the baseball is
\[ \lambda =\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )}= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )}=9.95\times 10^{-35}\; m \]
(You should verify that the units cancel to give the wavelength in meters.) Given that the diameter of the nucleus of an atom is approximately 10−14 m, the wavelength of the baseball is almost unimaginably small.
Find the de Broglie wavelength for an electron moving at the speed of \(5.0 \times 10^6\; m/s\) (mass of an electron is \(9.1 \times 10^{-31}\; kg\)).
- Answer:
-
This is a straightforward application of the de Broglie wavelength relation (Equation \ref{5}).
\[ \lambda = \dfrac{h}{p}= \dfrac{h}{mv} =\dfrac{6.63 \times 10^{-34}\; J \cdot s}{(9.1 \times 10^{-31} \; kg)(5.0 \times 10^6\, m/s)}= 1.46 \times 10^{-10}\;m = 1.46 Å \]
As you calculated in Example \(\PageIndex{1}\), objects such as a baseball have such short wavelengths that they are best regarded primarily as particles. In contrast, objects with very small masses (such as photons) have large wavelengths and can be viewed primarily as waves. Objects with intermediate masses, however, such as electrons (Example \(\PageIndex{2}\)), exhibit the properties of both particles and waves.To effectively diffract a wave, the wavelength of the wave must be comparable to the dimensions of the diffraction device!
Although we still usually think of electrons as particles, the wave nature of electrons is employed in an electron microscope, which has revealed most of what we know about the microscopic structure of living organisms and materials. Because the wavelength of an electron beam is much shorter than the wavelength of a beam of visible light, this instrument can resolve smaller details than a light microscope can:
![1479190a2b6f34e5b0992accfc3e979b.jpg](https://chem.libretexts.org/@api/deki/files/27819/1479190a2b6f34e5b0992accfc3e979b.jpg?revision=1&size=bestfit&width=709&height=281)
Standing Waves
De Broglie also investigated why only certain orbits were allowed in Bohr’s model of the hydrogen atom. He hypothesized that the electron behaves like a standing wave (a wave that does not travel in space). An example of a standing wave is the motion of a string of a violin or guitar. When the string is plucked, it vibrates at certain fixed frequencies because it is fastened at both ends (Figure \(\PageIndex{3}\)). If the length of the string is \(L\), then the lowest-energy vibration (the fundamental) has wavelength
\[ \begin{align} \dfrac{\lambda }{2} & =L \nonumber \\ \lambda &= 2L \nonumber \end{align} \label{6}\]
Higher-energy vibrations are called overtones (the vibration of a standing wave that is higher in energy than the fundamental vibration) and are produced when the string is plucked more strongly; they have wavelengths given by
\[ \lambda=\dfrac{2L}{n} \label{7}\]
where n has any integral value. When plucked, all other frequencies die out immediately. Only the resonant frequencies survive and are heard. Thus, we can think of the resonant frequencies of the string as being quantized. Notice in Figure \(\PageIndex{3}\) that all overtones have one or more nodes, points where the string does not move. The amplitude of the wave at a node is zero.
![d7d6077cda061c03591398212effebdf.jpg](https://chem.libretexts.org/@api/deki/files/27821/d7d6077cda061c03591398212effebdf.jpg?revision=1&size=bestfit&width=308&height=528)
Quantized vibrations and overtones containing nodes are not restricted to one-dimensional systems, such as strings. A two-dimensional surface, such as a drumhead, also has quantized vibrations. Similarly, when the ends of a string are joined to form a circle, the only allowed vibrations are those with wavelength
\[2πr = nλ \label{8}\]
where \(r\) is the radius of the circle. De Broglie argued that Bohr’s allowed orbits could be understood if the electron behaved like a standing circular wave (Figure \(\PageIndex{4}\)). The standing wave could exist only if the circumference of the circle was an integral multiple of the wavelength such that the propagated waves were all in phase, thereby increasing the net amplitudes and causing constructive interference. Otherwise, the propagated waves would be out of phase, resulting in a net decrease in amplitude and causing destructive interference. The nonresonant waves interfere with themselves! De Broglie’s idea explained Bohr’s allowed orbits and energy levels nicely: in the lowest energy level, corresponding to \(n = 1\) in Equation \(\ref{8}\), one complete wavelength would close the circle. Higher energy levels would have successively higher values of n with a corresponding number of nodes.
![210c78bb88b8b08f318dd62c87a3977a.jpg](https://chem.libretexts.org/@api/deki/files/27822/210c78bb88b8b08f318dd62c87a3977a.jpg?revision=1&size=bestfit&width=446&height=194)
Like all analogies, although the standing wave model helps us understand much about why Bohr's theory worked, it also, if pushed too far, can mislead. As you will see, some of de Broglie’s ideas are retained in the modern theory of the electronic structure of the atom: the wave behavior of the electron and the presence of nodes that increase in number as the energy level increases. Unfortunately, his (and Bohr's) explanation also contains one major feature that we now know to be incorrect: in the currently accepted model, the electron in a given orbit is not at the same distance from the nucleus.
Summary
An electron possesses both particle and wave properties. The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called wave–particle duality. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle.
\[\lambda =\dfrac{h}{mv} \nonumber \]
The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the fundamental vibration, and higher-energy vibrations are overtones and have successively more nodes, points where the amplitude of the wave is always zero.