Wave-particle duality: De Broglie and the photoelectric effect
A rooftop solar panel in Phoenix turns sunlight into electricity, yet in a physics lab a very bright red lamp can fail to free a single electron from a metal plate that a faint ultraviolet lamp discharges instantly. Why? We will come back to this at the end.
For two centuries, light and matter seemed to belong to different worlds: light was a wave, able to interfere and diffract; matter was made of particles with definite positions and trajectories. The photoelectric effect showed that light, as well as being a wave, behaves like a stream of particles. And in 1924, Louis de Broglie turned the idea around: if light can behave like a particle, perhaps matter can also behave like a wave. He was right, and this double nature of everything that exists is called wave-particle duality.
The photon: light as a particle
A heated metal bar glows red, then yellow, then white. Classical physics could not explain this glow, but it works if light energy comes in packets: why can light not carry any amount of energy?
The simulator uses an illustrative scale, not the real values for light. It lets you vary the wavelength and frequency independently, but for light in a vacuum their product is fixed ($c = \lambda f$): when you halve the wavelength, the frequency doubles, and the energy of each photon ($E=hf$) doubles with it. It is the same relationship you may have seen with emission spectra, now applied to the particle-like behavior of light.
A dermatological laser emits light with frequency $f = 4.5\times10^{14}\ \text{Hz}$. The energy of a photon is $E_{fotón} = h\times f$, where…
A dermatological laser emits light with frequency $f = 4.5\times10^{14}\ \text{Hz}$. The energy of a photon is $E_{fotón} = h\times f$, where $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.
- a) Calculate the energy of each photon.
- b) If we increase the intensity of the laser without changing its frequency, does the energy of each photon increase? What actually increases?
A laser pointer emits green light with wavelength $\lambda = 550\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and…
A laser pointer emits green light with wavelength $\lambda = 550\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $c = 3\times10^{8}\ \text{m/s}$.
- a) Calculate its frequency, $f = \frac{c}{\lambda}$.
- b) Calculate the energy of each photon, $E_{fotón} = h\times f$.
Sodium vapor street lamps emit light at $\lambda_{Na} = 589\ \text{nm}$ and a blue LED emits at $\lambda_{az} = 470\ \text{nm}$. Given:…
Sodium vapor street lamps emit light at $\lambda_{Na} = 589\ \text{nm}$ and a blue LED emits at $\lambda_{az} = 470\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $c = 3\times10^{8}\ \text{m/s}$. The diagram compares the two waves.
- a) Calculate the energy of a photon of each type of light.
- b) Which of the two photons is more energetic? How many times more?
The photoelectric effect
If we illuminate a negatively charged zinc plate with a very powerful beam of red light, it does not lose charge, but a faint ultraviolet lamp discharges it instantly: why does the frequency of the light determine the extraction of electrons, and not the total amount of light energy?
In the laboratory, a negatively charged zinc plate is connected to an electroscope. When illuminated with ultraviolet light, the electroscope discharges; with very…
In the laboratory, a negatively charged zinc plate is connected to an electroscope. When illuminated with ultraviolet light, the electroscope discharges; with very intense red light, it does not. The diagram shows the experimental setup.
- a) What is the photoelectric effect?
- b) What are photoelectrons?
- c) Explain, using the concept that light arrives in photons, why red light does not discharge the plate no matter how intense it is.
In the photoelectric effect, each metal has a work function $W$ (also called the extraction work) and a threshold frequency $f_0$ satisfying $W = h\times f_0$. a)…
In the photoelectric effect, each metal has a work function $W$ (also called the extraction work) and a threshold frequency $f_0$ satisfying $W = h\times f_0$.
- a) Explain what the work function means: what energy does it represent?
- b) What is the threshold frequency and what happens when the incident light has $f < f_0$?
- c) If $f > f_0$, what happens to the remaining photon energy after the electron is emitted?
A potassium photocell has a work function of $W = 2.3\ \text{eV}$ and is illuminated with ultraviolet light of $\lambda = 250\ \text{nm}$. Given:…
A potassium photocell has a work function of $W = 2.3\ \text{eV}$ and is illuminated with ultraviolet light of $\lambda = 250\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. The diagram shows the schematic of the process.
- a) Express the work function in joules.
- b) Calculate the energy of each photon of ultraviolet light.
- c) Apply $E_{fotón} = W + E_{c,máx}$ and calculate the maximum kinetic energy of the photoelectrons, in joules and in electronvolts.
De Broglie: matter as a wave
The double slit: the definitive proof of duality
When paintballs are fired at two slits, two bands of impacts are obtained, but when electrons are fired through two narrow slits, alternating fringes appear on the detection screen, with regions where none ever lands: how can individual material particles form a pattern identical to that of two overlapping waves?
Superposition of quantum states
How can a single electron build fringes? Quantum physics says that before it lands, the electron is not "at" slit 1 or "at" slit 2: its state is a superposition of both possibilities, like two overlapping waves. The two parts interfere, and that sets where the electron is likely to land. When it hits the screen, we always detect just one dot. If a detector checks which slit it used, the superposition disappears and so do the fringes. This is how superposition of quantum states connects to wave-particle duality: the wave aspect describes the possibilities, the particle aspect shows up when we detect it.
Light produces interference and diffraction patterns, like any wave, but when it hits a metal it knocks out electrons as if it were a stream of particles (photons). a)…
Light produces interference and diffraction patterns, like any wave, but when it hits a metal it knocks out electrons as if it were a stream of particles (photons).
- a) Explain in your own words what the wave-particle duality of light consists of.
- b) Name one phenomenon that can only be explained by wave behavior and another that can only be explained by particle behavior.
- c) In $1924$, De Broglie extended the idea to matter: what hypothesis did he propose regarding electrons and other particles?
Stopping potential and photon flux
A laboratory group measures the photoelectric effect on a metal and obtains two data points: with light of frequency $f_1 = 6.0\times10^{14}\ \text{Hz}$ there is no…
A laboratory group measures the photoelectric effect on a metal and obtains two data points: with light of frequency $f_1 = 6.0\times10^{14}\ \text{Hz}$ there is no emission at all (it is the threshold frequency) and with $f_2 = 1.0\times10^{15}\ \text{Hz}$ photoelectrons leave with $E_{c,máx} = 2.65\times10^{-19}\ \text{J}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. The diagram shows the two data points from the experiment.
- a) Calculate the work function of the metal from $W = h\times f_1$ and express it in electronvolts.
- b) Mentally sketch $E_{c,máx}$ against $f$: using $E_{c,máx} = h\times f - W$, calculate the slope of the straight line through the two points and check that it matches the value of $h$.
- c) Predict, without measuring again, the maximum kinetic energy for light of $f = 1.2\times10^{15}\ \text{Hz}$.
- d) Explain what quantity the y-intercept of the straight line represents and why it is negative.
A zinc plate, with a work function of $W = 4.3\ \text{eV}$, is illuminated alternately with red light of $650\ \text{nm}$ and ultraviolet light of $250\ \text{nm}$.…
A zinc plate, with a work function of $W = 4.3\ \text{eV}$, is illuminated alternately with red light of $650\ \text{nm}$ and ultraviolet light of $250\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.
- a) Calculate the energy of each type of photon and determine, with justification, which light causes emission.
- b) For the light that produces emission, calculate the maximum kinetic energy of the photoelectrons and the corresponding stopping potential.
- c) If we multiply the intensity of the red light by a thousand, will we succeed in knocking out electrons from the zinc? Explain your answer using the photon model: how do a thousand red photons differ from one ultraviolet photon?
- d) If we increase the intensity of the ultraviolet light, what changes and what stays the same in the emission?
A particle accelerated from rest through a potential difference gains, without losses, $KE = |q\Delta V|$. For non-relativistic electrons, combine this with $p=\sqrt{2m\,KE}$ and $\lambda=h/p$. Check that $KE$ is much smaller than $mc^2$ before using this approximation.
The diagram links the main ideas:
A fiber-optic laser for telecommunications emits at $\lambda = 1.55\ \mu\text{m}$ with a power of $10\ \text{mW}$. Given:…
A fiber-optic laser for telecommunications emits at $\lambda = 1.55\ \mu\text{m}$ with a power of $10\ \text{mW}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $c = 3\times10^{8}\ \text{m/s}$.
- a) Calculate the energy of each photon.
- b) Calculate how many photons the laser emits each second (remember that power is the energy emitted per unit time).
A metal with work function $W = 2.25\ \text{eV}$ is illuminated with light of $\lambda = 300\ \text{nm}$. The photoelectrons are slowed down by a potential difference…
A metal with work function $W = 2.25\ \text{eV}$ is illuminated with light of $\lambda = 300\ \text{nm}$. The photoelectrons are slowed down by a potential difference until even the fastest one does not reach the collector (stopping potential $V_f$, with $E_{c,máx} = e\times V_f$). Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$, $e = 1.6\times10^{-19}\ \text{C}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. The diagram shows the setup with the stopping potential.
- a) Calculate the photon energy and the maximum kinetic energy of the photoelectrons, in electronvolts.
- b) Calculate the stopping potential.
- c) Repeat the calculation with light of $\lambda = 250\ \text{nm}$: how do the maximum kinetic energy and the stopping potential change?
- d) Explain why doubling the light intensity, without changing its wavelength, does not change the stopping potential, even though it increases the number of photoelectrons.
A photovoltaic cell receives $5\ \text{mW}$ of light of $\lambda = 600\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$,…
A photovoltaic cell receives $5\ \text{mW}$ of light of $\lambda = 600\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$ and $e = 1.6\times10^{-19}\ \text{C}$.
- a) Calculate how many photons reach the cell each second.
- b) Assuming that each photon releases one electron that reaches the external circuit, calculate the maximum theoretical current (current is charge passing per second).
- c) In practice the current is lower: give two reasons to explain this difference.
- d) If the cell received the same power but from light of $\lambda = 400\ \text{nm}$, would the maximum theoretical current be greater, smaller, or equal? Justify your answer by counting photons.
Back to the start: a red lamp, however bright, delivers photons that each carry too little energy to free an electron, while each ultraviolet photon carries enough. A solar panel works because sunlight contains photons with enough energy, and brighter sunlight means more photons per second, so more current.