Wave-particle duality: De Broglie and the photoelectric effect
The street lights on your road switch on by themselves at dusk: a small photocell detects the fading daylight. How can light, which is a wave, push electrons out of a metal as if it were a stream of tiny particles? 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
Classical wave theory could not explain the photoelectric effect. Einstein proposed that light transfers its energy in packets (photons).
The simulator lets you vary the wavelength and frequency independently. In a given medium $\lambda f = c$ is fixed, so halving $\lambda$ doubles $f$ and therefore doubles the energy of each photon ($E=hf$). It is the same relationship that links the energy levels of an atom to its emission spectrum ($E_2 - E_1 = hf$), now applied to the particle-like behaviour 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?
A photocell works exactly this way: light of high enough frequency knocks electrons out of a metal surface, and these electrons form a current that a circuit can detect. This is how the street lights on your road know when to switch on at dusk.
Real applications of photocells include automatic doors, light meters, burglar alarms and the light sensors in cameras. In each one, the current rises with the number of photons arriving per second, as long as $f > f_0$. Note the link with X-rays, which are the reverse process. In the photoelectric effect a photon is absorbed and an electron is freed; in an X-ray tube, fast electrons accelerated through a high p.d. hit a metal target and some of their kinetic energy is emitted as high-frequency photons. In both cases energy is conserved and the photon energy is $E = hf$. The highest-energy X-ray photon has $hf = E_k$ of the incoming electron. Doubling the intensity at fixed $f>f_0$ doubles the number of photons per second, so the photocurrent is proportional to the intensity (in the ideal case where each photon releases one electron that reaches the circuit), but $E_{k,max}$ is unchanged.
The effect supports the particle model of light: wave theory cannot explain why emission depends on frequency and not on intensity, whereas the photon model explains it directly. You can verify this with secondary data: plot the maximum kinetic energy of the photoelectrons against frequency for a metal. You get a straight line with gradient $h$ that cuts the frequency axis at $f_0$, whatever the intensity. In the Leaving Certificate specification the work function is written $\Phi$ ($\Phi = hf_0$); here we write $W$, as in the exercises, so that it is not confused with magnetic flux.
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 set-up.
- 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.
De Broglie: matter as a wave
Every particle has an associated De Broglie wavelength, $\lambda = \frac{h}{m\times v}$, where $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$. A tennis ball of mass…
Every particle has an associated De Broglie wavelength, $\lambda = \frac{h}{m\times v}$, where $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$. A tennis ball of mass $58\ \text{g}$ leaves a serve at $40\ \text{m/s}$.
- a) Calculate its De Broglie wavelength.
- b) Compare the result with the size of an atom ($10^{-10}\ \text{m}$) and explain why this wave has no observable consequences.
In a cathode-ray tube, electrons move at $v = 10^{6}\ \text{m/s}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and mass of an electron…
In a cathode-ray tube, electrons move at $v = 10^{6}\ \text{m/s}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and mass of an electron $m_e = 9.11\times10^{-31}\ \text{kg}$.
- a) Calculate the De Broglie wavelength associated with each electron, $\lambda = \frac{h}{m_e\times v}$.
- b) Compare the result with the typical size of an atom ($10^{-10}\ \text{m}$) and comment on why the wave associated with the electron is relevant in the atomic world.
In an accelerator, an electron and a proton move at the same speed, $v = 10^{5}\ \text{m/s}$. Given: $m_e = 9.11\times10^{-31}\ \text{kg}$,…
In an accelerator, an electron and a proton move at the same speed, $v = 10^{5}\ \text{m/s}$. Given: $m_e = 9.11\times10^{-31}\ \text{kg}$, $m_p = 1.67\times10^{-27}\ \text{kg}$ and $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.
- a) Calculate the De Broglie wavelength of the electron.
- b) Calculate that of the proton.
- c) Calculate the ratio $\frac{\lambda_e}{\lambda_p}$ and explain, using De Broglie's formula, why at the same speed the more massive particle has a shorter associated wavelength.
In a transmission electron microscope, electrons are accelerated through a potential difference $V = 10\ \text{kV}$, gaining a kinetic energy $E_c = e\times V$ (ignore…
In a transmission electron microscope, electrons are accelerated through a potential difference $V = 10\ \text{kV}$, gaining a kinetic energy $E_c = e\times V$ (ignore relativistic corrections). Given: $e = 1.6\times10^{-19}\ \text{C}$, $m_e = 9.11\times10^{-31}\ \text{kg}$ and $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.
- a) Calculate the kinetic energy acquired by each electron, in joules.
- b) Calculate its speed.
- c) Calculate its De Broglie wavelength.
- d) The resolving power of a microscope improves as the wavelength used decreases: compare the wavelength obtained with that of visible light ($\lambda \approx 500\ \text{nm}$) and explain why an electron microscope can resolve viruses while an optical microscope cannot.
The double slit: the definitive proof of duality
When paint pellets 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?
For a certain diffraction experiment, the electrons must have a De Broglie wavelength of $1\ \text{nm}$. Given: $m_e = 9.11\times10^{-31}\ \text{kg}$,…
For a certain diffraction experiment, the electrons must have a De Broglie wavelength of $1\ \text{nm}$. Given: $m_e = 9.11\times10^{-31}\ \text{kg}$, $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.
- a) Rearrange $\lambda = \frac{h}{m_e\times v}$ for speed and calculate the speed at which they must move.
- b) Calculate their kinetic energy, in joules and in electronvolts.
In a double-slit experiment with electrons, the electrons move at $v = 2\times10^{6}\ \text{m/s}$, the slits are separated by $d = 0.5\ \mu\text{m}$, and the screen is…
In a double-slit experiment with electrons, the electrons move at $v = 2\times10^{6}\ \text{m/s}$, the slits are separated by $d = 0.5\ \mu\text{m}$, and the screen is at $L = 1\ \text{m}$. The separation between consecutive maxima in the pattern is $\Delta y = \frac{\lambda\times L}{d}$. Given: $m_e = 9.11\times10^{-31}\ \text{kg}$ and $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$. The diagram shows the double-slit set-up.
- a) Calculate the De Broglie wavelength of the electrons.
- b) Calculate the separation between maxima on the screen and express it in millimetres.
- c) If electrons are sent one by one, the interference pattern still forms over time: explain what this result means regarding wave-particle duality and the probabilistic nature of quantum physics.
- d) Predict what happens to the separation between maxima if the electrons are accelerated to twice the speed.
Stopping potential and photon flux
A particle accelerated from rest through a potential difference gains, without losses, $E_c=|q\Delta V|$ (that is, $qV=\tfrac12mv^2$, with $e=1.602\times10^{-19}\ \text{C}$). For non-relativistic electrons, combine this with $p=\sqrt{2mE_c}$ and $\lambda=h/p$. Check that $E_c$ is much smaller than $mc^2$ before using this approximation.
The diagram links the main ideas:
Back to the street lights: the metal in the photocell releases electrons only when each photon has enough energy, so the photons in daylight that are above the threshold frequency produce a current. At dusk fewer photons arrive, so the photocurrent falls. The circuit switches the lamp on when the current drops below a set value.
A fibre-optic laser for telecommunications emits at $\lambda = 1.55\ \mu\text{m}$ with a power of $10\ \text{mW}$. Given:…
A fibre-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 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.