Dual Nature of Radiation and Matter: common doubts, answered
The questions students ask most often about Dual Nature of Radiation and Matter, each with a short answer. For the full chapter, read the Dual Nature of Radiation and Matter notes.
About the chapter
What is meant by the dual nature of radiation and matter?
It means both light and matter show wave behaviour in some experiments and particle behaviour in others. Light interferes and diffracts like a wave, yet in the photoelectric effect it delivers energy in single photons. Electrons arrive at detectors as particles, yet they have a wavelength λ = h/p. Which description to use depends on the experiment; neither alone explains everything.
Electron emission
Read this section in the notes →What is the work function of a metal?
It is the least energy that must be given to an electron to free it from the metal's surface, written φ₀ and usually quoted in electron volts. Free electrons inside a metal cannot simply leave, because an escaping electron is pulled back by the positive ions it leaves behind. The work function depends on the metal and on the condition of its surface; a low value means electrons escape more easily.
In what ways can electrons be emitted from a metal surface?
There are three ways: thermionic emission, field emission and photoelectric emission. Heating the metal gives free electrons enough thermal energy to escape; a very strong electric field, of the order of 10⁸ V m⁻¹, can pull them out, as happens in a spark plug; and light of a suitable frequency can knock them out. In every case each electron must gain at least the work function.
What is an electron volt and how is it converted to joules?
An electron volt is the energy an electron gains when it is accelerated through a potential difference of one volt: 1 eV = 1.6 × 10⁻¹⁹ J. It is a handy size for the energies of electrons and photons. Before using Kmax = hν − φ₀, put every term in the same unit, because mixing photon energy in joules with a work function in electron volts is a very common error.
Discovery of the photoelectric effect
Read this section in the notes →Who discovered the photoelectric effect and how was it first studied?
Heinrich Hertz first noticed it, finding that ultraviolet light on the electrodes made sparks jump more easily. Hallwachs and Lenard then studied it carefully. Hallwachs saw a negatively charged zinc plate lose its charge under ultraviolet light, showing that negative particles were being driven out, and Lenard found that light on a plate in an evacuated tube made a current flow that stopped when the light was cut off.
Saturation current and stopping potential
Read this section in the notes →What is saturation current in the photoelectric effect?
It is the largest photocurrent, reached when the collector is positive enough to gather every electron the emitter gives off. Raising the accelerating voltage beyond this point adds nothing, because the number of electrons emitted each second is already fixed by the light. The saturation current is proportional to the intensity of the incident light, since intensity sets how many photoelectrons are released.
What is stopping potential in the photoelectric effect?
It is the smallest negative potential of the collector, relative to the emitter, that just stops the photocurrent. At that point even the fastest photoelectrons turn back before reaching the collector, so Kmax = eV₀. It is a retarding potential, so the collector is negative, not positive. Its value depends on the frequency of the light and on the metal, but not on the intensity.
Why does brighter light increase the photocurrent but not the stopping potential?
Because brighter light of the same frequency means more photons each second, not more energetic ones. More photons free more electrons, so the saturation current rises in proportion to intensity. Each electron still receives the energy of a single photon, hν, so the fastest electrons have the same Kmax as before, and the stopping potential, given by Kmax = eV₀, does not change.
Threshold frequency and no time lag
Read this section in the notes →Why can't very bright red light eject electrons from a metal whose threshold lies in the ultraviolet?
Because each electron absorbs one photon at a time, and a red photon carries less energy than the work function of such a metal. Brighter light only brings more of these weak photons, and their energies cannot be pooled into one electron. So below the threshold frequency ν₀ = φ₀/h there is no emission at all, however intense the light.
Why is there no time lag in photoelectric emission even in dim light?
Because each emission is a single event in which one photon gives all its energy to one electron, and that happens at once. Even in very dim light some photons arrive immediately, and each can eject an electron straight away. Dim light simply means fewer electrons per second. The wave picture predicted a delay while energy spread thinly over the surface built up in an electron.
Where the wave picture fails
Read this section in the notes →Why couldn't the wave theory of light explain the photoelectric effect?
Because the wave picture links a wave's energy to its intensity. It therefore predicted that brighter light should give faster electrons, that any frequency should work if the light were intense enough, and that dim light would need time to build up enough energy. Experiment contradicted all three: Kmax depends on frequency alone, there is a threshold frequency, and emission is instantaneous.
Einstein's photoelectric equation
Read this section in the notes →What does Einstein's photoelectric equation mean physically?
Kmax = hν − φ₀ says a photon's energy hν first pays the cost of freeing an electron, at least the work function φ₀, and whatever remains becomes the electron's kinetic energy. The least tightly bound electrons keep all the remainder and leave with Kmax; others lose more energy getting out and emerge slower. The equation holds for ν ≥ ν₀, and with Kmax = eV₀ it links stopping potential to frequency.
Why is the slope of the stopping potential versus frequency graph the same for every metal?
Because the graph follows V₀ = (h/e)ν − φ₀/e, and its slope h/e contains only universal constants. Changing the metal changes only the work function, which moves the line sideways and changes where it meets the frequency axis, at ν₀ = φ₀/h. So the lines for different metals are parallel. Millikan used this slope, with the known e, to find a value of Planck's constant.
The photon
Read this section in the notes →What is a photon?
A photon is a quantum, or packet, of electromagnetic energy, with energy E = hν and momentum p = hν/c = h/λ. It travels at the speed of light in vacuum. All photons of a given frequency carry the same energy and momentum, whatever the intensity of the light. A more intense beam of the same colour simply carries more photons across a given area each second.
How do I find the number of photons emitted per second by a light source?
Divide the power of the beam by the energy of one photon: N = P/E, with E = hν = hc/λ. For example, a source giving out 1 W of light at a wavelength where each photon carries 3 × 10⁻¹⁹ J emits about 3.3 × 10¹⁸ photons every second. For the same power, a longer wavelength means less energy per photon and therefore more photons.
Can a photon be deflected by electric or magnetic fields?
No. A photon carries no electric charge, so electric and magnetic fields exert no force on it and do not bend its path. In a collision between a photon and a particle such as an electron, the total energy and total momentum are conserved. The number of photons, however, need not stay the same, because a photon can be absorbed or a new one created.
Matter waves and de Broglie
Read this section in the notes →What is the de Broglie wavelength of a particle?
It is the wavelength linked with a moving particle: λ = h/p = h/mv. De Broglie argued that since light, a wave, also behaves like particles, matter, made of particles, should also behave like waves, using the same relation between momentum and wavelength as for photons. The wavelength depends only on momentum; the particle's charge and what it is made of play no part.
Does a heavier or faster particle have a longer de Broglie wavelength?
No, a shorter one. Since λ = h/mv, the wavelength falls as the momentum rises, so doubling the mass or the speed halves it. A proton and an electron moving at the same speed have very different wavelengths, the proton's being much shorter. If instead they have equal momentum, their wavelengths are equal, despite their very different masses.
How does the de Broglie wavelength of an electron depend on its kinetic energy or accelerating voltage?
It varies inversely as the square root of either. Momentum and kinetic energy K are related by p = √(2mK), so λ = h/√(2mK). For an electron accelerated from rest through a potential difference V, K = eV, giving λ = h/√(2meV). Making the voltage four times larger therefore halves the wavelength. These relations hold for speeds well below the speed of light.
How long is a matter wave?
Read this section in the notes →Why don't we see the wave nature of everyday objects like a cricket ball?
Because their de Broglie wavelengths are far too small to detect. A ball of 0.15 kg moving at 30 m/s has momentum 4.5 kg m/s, giving λ = h/p of about 1.5 × 10⁻³⁴ m, vastly smaller than even a proton. No instrument can measure such a length. An electron's tiny mass gives it a wavelength comparable to the spacing of atoms in a crystal, so its wave behaviour can be observed.
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