Dual Nature of Radiation and Matter

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 11.

Why does a faint UV lamp beat a bright red floodlight?

Shine a weak ultraviolet lamp on a clean zinc plate and electrons leave it at once. Shine a red floodlight a million times brighter on the same plate, for hours, and nothing happens. Classical physics says this is absurd: more energy should always win. This chapter is about why it does not, and the answer changed physics. Light does not hand over its energy in a smooth stream. It arrives in packets, and a packet is either big enough or it is not.

The second half of the chapter flips the idea around. If light waves can behave like particles, can particles such as electrons behave like waves? They can, and that is why an electron microscope exists.

Getting an electron out of a metal

Inside a metal, conduction electrons wander freely, but they cannot simply walk out. The positive ions just inside the surface pull them back, so an electron must do work to escape. The least work it needs is the work function $\phi_0$ of that metal. It is measured in electron volts, where $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$ is the energy an electron gains by moving through a potential difference of 1 volt.

The work function belongs to the metal and its surface, not to the light. Caesium has the smallest common value, about 2.14 eV. Sodium is about 2.75 eV, copper about 4.7 eV and platinum about 5.65 eV. A dirty or oxidised surface changes the number, which is why the experiments below need a freshly cleaned, evacuated tube.

There are three ways to supply this energy:

How the photoelectric effect was found

In 1887 Heinrich Hertz noticed that sparks across his detector gap were stronger when ultraviolet light fell on the emitting plate. In 1888 Wilhelm Hallwachs showed what was escaping. A freshly polished, negatively charged zinc plate lost its charge under UV light, an uncharged plate became positive, and a positive plate became more positive. Only negative charges leaving the plate explain all three. Philipp Lenard then built the sealed, evacuated tube that is still used today.

The apparatus in words. An evacuated quartz or glass tube holds two metal plates: the emitter C, which is lit, and the collector A. Monochromatic light enters through a quartz window, because ordinary glass blocks UV. A battery and a reversing switch let you make A positive or negative relative to C. A voltmeter reads that potential difference V and a microammeter reads the photocurrent.

What the experiment shows

Vary one thing at a time and watch the current. Everything below assumes the light's frequency is above the metal's threshold.

Intensity. Make the light brighter at fixed frequency and fixed positive V. The photocurrent rises in direct proportion to intensity. More light means more electrons per second.

Collector potential. Make A more positive and the current climbs, then flattens at the saturation current once every ejected electron reaches the collector. Now reverse the battery. The current falls because the field slows the electrons, and at one sharp negative value, the stopping potential $V_0$, it becomes exactly zero. At that point even the fastest electron is turned back just before A, so $K_{\max} = eV_0$. Double the intensity and the saturation current doubles, but $V_0$ stays exactly where it was.

Frequency. Keep intensity fixed and raise the frequency. Saturation current barely changes, but $V_0$ grows steadily. Plot $V_0$ against $\nu$ and you get a straight line. It cuts the frequency axis at the threshold frequency $\nu_0$. Below $\nu_0$ nothing is emitted, however bright the light or however long you wait.

Time. Emission starts in less than about $10^{-9}$ s after the light is switched on, even for extremely dim light.

Why the wave picture fails

Classical wave theory spreads energy evenly over the whole wavefront. That makes four predictions, and the experiment contradicts all four.

Einstein's photon explanation

In 1905 Einstein proposed that light of frequency $\nu$ is made of packets called photons, each carrying energy $h\nu$, where $h$ is Planck's constant. A photoelectron is produced by a single photon giving all its energy to a single electron. Part of that energy, $\phi_0$, pays the cost of leaving the metal. The rest becomes kinetic energy.

$$K_{\max} = h\nu - \phi_0 = eV_0$$

This is just conservation of energy for one absorption event. Every observation follows from it.

Intensity now means the number of photons per second, not the energy of each photon. More photons per second knock out more electrons per second, so the current rises, but each electron still gets $h\nu$, so $K_{\max}$ does not change. If $h\nu < \phi_0$, no single photon can free an electron, and a million of them arriving separately make no difference, because the electron cannot save up energy from several photons. That is the threshold. Setting $K_{\max} = 0$ gives it directly:

$$\nu_0 = \dfrac{\phi_0}{h} \quad \text{and} \quad \lambda_0 = \dfrac{c}{\nu_0} = \dfrac{hc}{\phi_0}$$

A light packet is absorbed in one go, so there is no waiting time. Rearranging Einstein's equation gives the line you plot in the lab:

$$V_0 = \dfrac{h}{e}\,\nu - \dfrac{\phi_0}{e}$$

Its slope is $h/e$, a universal constant, so every metal gives a parallel line. Only the intercept on the frequency axis, $\nu_0$, changes from metal to metal. Robert Millikan set out to disprove Einstein but spent years measuring this slope for alkali metals and ended up confirming it, which gave an accurate value of $h \approx 6.63 \times 10^{-34} \text{ J s}$. Einstein's 1921 Nobel Prize was for this law, not for relativity.

A useful shortcut: $hc = 1240 \text{ eV nm}$, so a photon of wavelength $\lambda$ in nm has energy $1240/\lambda$ in eV. This lets you do most CUET numericals without touching powers of ten.

Why Kmax is a maximum. The equation gives the energy of the fastest electron. An electron buried deeper than the surface layer loses extra energy getting out, so it emerges slower. That is why the stopping potential, which stops even the fastest, measures Kmax.

Worked example: stopping potential and threshold. Light of wavelength 400 nm falls on a metal with $\phi_0 = 2.0 \text{ eV}$. Find $K_{\max}$, $V_0$, the threshold wavelength and the fastest electron's speed.

The common wrong step: doubling the intensity and claiming $V_0$ doubles. It does not. Doubling intensity doubles the number of photons, so the saturation current doubles, while each photon still delivers 3.10 eV. Light of wavelength above 620 nm, such as red light at 700 nm, will not eject electrons at any intensity.

Beyond the textbook. For the exam: no emission below $\nu_0$, however intense the light, because one photon meets one electron. In reality: this holds for ordinary light. With the extremely intense pulses of a modern laser, an electron can absorb two or more photons almost at once (multiphoton photoemission), so photoelectrons can appear even when a single photon is below the threshold. The one photon, one electron picture is the rule to use in every school and CUET problem. Why it matters: the threshold is a statement about single absorptions, not about energy in general.

Pause and predict: a metal has $\nu_0 = 5 \times 10^{14}$ Hz. You shine light of $4 \times 10^{14}$ Hz and then double its intensity three times. What is the saturation current?

Check your answer: zero. Each photon carries $h \times 4 \times 10^{14}$ Hz, which is less than $\phi_0 = h\nu_0$. Increasing the number of photons does not help, because no single photon can free an electron.

Worked example: finding $h$ and $\phi_0$ from a graph. In a photoelectric experiment the stopping potential is 0.40 V for $\nu = 6.0 \times 10^{14}$ Hz and 1.64 V for $\nu = 9.0 \times 10^{14}$ Hz. Find $h$, $\phi_0$ and $\nu_0$. Take $e = 1.6 \times 10^{-19} \text{ C}$.

The common wrong step: dividing $V_0$ by $\nu$ at one point and calling the answer $h/e$. That would be right only if the line passed through the origin, which it does not, because of the $\phi_0/e$ intercept. Always use the slope between two points.

The photon: a particle of light

A photon has energy $E = h\nu = hc/\lambda$, momentum $p = h\nu/c = h/\lambda$ and speed $c$. It carries no charge, so electric and magnetic fields do not deflect it. Its rest mass is zero, and for something with zero rest mass the relation is $E = pc$. In a collision between a photon and a particle, energy and momentum are conserved, but the number of photons is not: a photon can be absorbed or a new one created.

Intensity of a beam is the energy crossing unit area per second, which equals (photons per second per unit area) $\times h\nu$. Two beams of different colours with the same intensity therefore have different photon counts: the redder beam has more photons.

Worked example: counting photons. A laser pointer emits 5.0 mW at 633 nm. How many photons leave it each second, and what is the momentum of one?

The common wrong step: leaving the power as 5.0 mW and getting an answer a thousand times too large, or putting $\lambda = 633$ into $p = h/\lambda$ without converting from nm to metres.

If light can be a particle, can matter be a wave?

In 1924 Louis de Broglie reasoned that nature likes symmetry. A photon has $p = h/\lambda$, so any moving particle of momentum $p$ should have an associated wave of wavelength

$$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$$

This is the de Broglie wavelength. It is consistent with light, since for a photon $\lambda = h/p$ becomes $\lambda = c/\nu$. Why have we never seen a car interfere with itself? Because for any large object $\lambda$ is absurdly small. A 0.12 kg cricket ball at 20 m s⁻¹ has $\lambda = 6.63 \times 10^{-34}/(0.12 \times 20) \approx 2.8 \times 10^{-34} \text{ m}$, far below anything that could be detected, and the same is true of a 150 g ball at 30 m s⁻¹, where $\lambda \approx 1.5 \times 10^{-34} \text{ m}$. Matter waves only matter when the mass is tiny.

Electrons accelerated through a potential difference. An electron starting from rest and accelerated through V volts gains kinetic energy $eV = p^2/2m$, so $p = \sqrt{2meV}$ and

$$\lambda = \dfrac{h}{\sqrt{2meV}} \approx \dfrac{12.27}{\sqrt{V}} \text{ Å}$$

The Å version works only for an electron, with V in volts. For other charged particles use $\lambda = h/\sqrt{2mqV}$. Putting in the numbers gives these handy forms:

An electron at 100 V has $\lambda \approx 1.23 \text{ Å}$, about the spacing between atoms in a crystal. That is the clue that a crystal should diffract it.

Davisson and Germer, 1927. They fired electrons accelerated through 54 V at a nickel crystal and found a sharp peak in the scattered intensity at 50° to the incident beam. Treating the crystal planes like a grating (Bragg's law) gave a wavelength of 1.65 Å. The de Broglie formula gave $12.27/\sqrt{54} = 1.67 \text{ Å}$. The match confirmed that electrons diffract like waves. G. P. Thomson independently saw electron diffraction in thin films the same year.

Worked example: an electron through 150 V. Find its de Broglie wavelength, and say what happens if the voltage is raised to 600 V.

The common wrong step: writing $\lambda = 12.27/V$ and getting 0.082 Å. The voltage appears under a square root because $\lambda$ depends on momentum, and momentum goes as the square root of energy.

Worked example: same wavelength, different particles. A proton accelerated through potential $V$ has de Broglie wavelength $\lambda$. Through what potential must an α-particle be accelerated to have the same $\lambda$?

The common wrong step: using only the charge ratio (giving $V/2$) or only the mass ratio (giving $V/4$). Both mass and charge sit under the square root, so their product matters.

Pause and predict: an electron and a proton have the same de Broglie wavelength. Which one has more kinetic energy?

Check your answer: the electron. Equal $\lambda$ means equal momentum $p$, and $K = p^2/2m$. For the same $p$, the lighter particle has the greater kinetic energy, by a factor of about 1836.

Beyond the textbook. For the exam: an electron accelerated through $V$ volts has $\lambda = 12.27/\sqrt{V}$ Å. In reality: this formula ignores relativity. In an electron microscope at 100 kV the electrons move at about half the speed of light, and the relativistic correction changes $\lambda$ by a few per cent, to about 3.7 pm (0.0037 nm). That is tens of thousands of times shorter than visible light, yet real microscopes do not reach that resolution, because lens imperfections, not the wavelength, set the limit. Why it matters: wavelength is the theoretical resolution limit, but good lenses are what let you approach it.

Light is both wave and particle. Light spreads, diffracts and interferes when it travels, and arrives in lumps when it is absorbed. These are two faces of one thing, not two competing theories, and the same is true for electrons. Which face you see depends on the experiment you do.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Wave Optics shows the interference and diffraction that make light a wave. This chapter shows the other side of the same light.
• Next: Atoms uses photon energy for the hydrogen spectrum and uses de Broglie's idea to justify Bohr's quantised orbits.
• Related: Electromagnetic Waves places UV, visible and X-ray photons on one spectrum, and Electrostatic Potential and Capacitance is where the electron volt and the energy of a charge in a potential difference come from.