Dual Nature of Radiation and Matter

By the end of the nineteenth century, two things looked settled. Light was an electromagnetic wave: Maxwell had predicted it, and in 1887 Hertz produced and detected radio waves, confirming the picture. Matter was made of particles, and one of them had just been found. Studies of electric discharge through gases at low pressure had shown that cathode rays were streams of negatively charged particles, and in 1897 J. J. Thomson measured their charge-to-mass ratio, e/m = 1.76 × 10¹¹ C kg⁻¹. He found the same value whatever the gas or the cathode material, which meant the electron is a universal constituent of matter. Millikan's oil-drop experiment (1913) later gave the charge itself, e = 1.602 × 10⁻¹⁹ C, and showed that charge is quantised.

The chapter is about how this tidy division broke down. A metal contains free electrons, but they don't leak out on their own; something must supply energy to pull them across the surface. When light does the pulling, the details of what is seen cannot be explained by a wave picture of light. Einstein's explanation (1905) needed light to arrive in discrete packets, the photons. Then in 1924 de Broglie proposed the reverse: if radiation has a particle side, matter should have a wave side, and electrons should have a wavelength. Both halves were confirmed by experiment. So radiation and matter each have a dual nature, and the chapter treats the two halves in turn: electron emission and the photoelectric effect first, matter waves last.

For the exam, keep the constants you will use in every numerical at hand: h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, e = 1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg, and the shortcut hc ≈ 1240 eV nm.

Metals conduct because some of their electrons are free to move about inside the metal. Those electrons cannot normally escape, because as soon as one tries to leave, the positive ions it leaves behind pull it back. The surface acts as a barrier. The minimum energy needed by an electron to escape from the metal surface is called the work function of the metal, written φ₀. It depends on the metal and on the condition of its surface. Work functions are quoted in electron volts: 1 eV is the energy gained by an electron accelerated through 1 volt, so 1 eV = 1.602 × 10⁻¹⁹ J. The work function is different for every metal — NCERT's own worked example uses caesium's value, 2.14 eV. Caesium and the other alkali metals (lithium, sodium, potassium, rubidium) have such low work functions that even visible light can free their electrons, which is why they are used in photosensitive surfaces.

The energy to cross the barrier can be supplied in three ways, and the examiner expects the three names.

• Thermionic emission: heat the metal. The thermal energy of the free electrons rises and some acquire enough to leave. This is how the filament in a vacuum tube or an X-ray tube supplies electrons.
• Field emission: apply a very strong electric field, of the order of 10⁸ V m⁻¹, to the metal. The field pulls electrons out of the surface.
• Photoelectric emission: shine light of suitable frequency on the metal. Electrons come out, and they are called photoelectrons. This is the mechanism the rest of the chapter is about.

Note the word 'minimum' in the definition of work function. An electron deep inside the metal may need more than φ₀ to get out because it loses energy in collisions on the way. That is why photoelectrons emerge with a range of kinetic energies up to a maximum, and why Einstein's equation is written for K_max, not for every electron.

The photoelectric effect was found by accident. In 1887, while generating electromagnetic waves with a spark gap, Hertz noticed that the spark across the detector gap jumped more readily when ultraviolet light from the emitter spark fell on the metal electrodes. Light falling on the metal was somehow helping electrons out of it.

Hallwachs (1888) made the effect explicit. He connected a zinc plate to an electroscope. A negatively charged zinc plate lost its charge when ultraviolet light fell on it. An uncharged plate became positively charged, and a positively charged plate became still more positive under the same light. All three fit one idea: the ultraviolet light was ejecting negatively charged particles from the zinc.

Lenard (1902) built the experiment that the rest of the chapter uses. Two metal plates were sealed in an evacuated glass tube and a potential difference was applied between them. When ultraviolet light fell on the emitter plate, a current flowed in the external circuit; as soon as the light was cut off, the current stopped. So electrons were being emitted from the emitter and collected by the other plate. Hallwachs and Lenard also found that no electrons at all came out when the frequency of the light was below a certain minimum, however intense the light was. That minimum is the threshold frequency ν₀, and it depends on the metal. Alkali metals like lithium, sodium, potassium and caesium respond to visible light; zinc, cadmium and magnesium respond only to ultraviolet.

Put together, these observations define the effect: the emission of electrons from a metal surface when light of sufficiently high frequency falls on it. The electrons are photoelectrons and the current they carry is the photoelectric current.

The set-up is Lenard's tube drawn properly (Fig. 11.1 in NCERT). An evacuated glass or quartz tube holds a photosensitive emitter plate C and a collector plate A. Monochromatic light from a source S enters through a quartz window (glass absorbs ultraviolet) and falls on C. A battery and a potential divider let you make A positive or negative with respect to C and vary the potential; a voltmeter reads it and a microammeter reads the photocurrent. You can change three things independently: the intensity of the light, its frequency, and the collector potential. Three sets of observations follow.

(1) Effect of intensity. Keep frequency and collector potential fixed (collector positive) and vary intensity. The photocurrent is directly proportional to intensity: the graph of photocurrent against intensity is a straight line through the origin. Since photocurrent measures the number of photoelectrons emitted per second, this number is proportional to intensity.

(2) Effect of potential. Keep intensity and frequency fixed. As the collector is made more positive the current rises and then levels off at the saturation current, the value at which every emitted electron is being collected. Now make the collector negative. The current falls, and at a certain negative potential it becomes zero. This is the stopping potential (cut-off potential) V₀: the minimum retarding potential at which the photocurrent is zero. Even the fastest electrons are just turned back, so K_max = ½ m v_max² = eV₀. Repeating with a higher intensity of the same frequency, the saturation current is higher but V₀ is exactly the same. So stopping potential, and hence K_max, is independent of intensity.

(3) Effect of frequency. Keep intensity fixed and repeat at different frequencies. A higher frequency gives a higher stopping potential, though the saturation current is the same for the same intensity. Plotting V₀ against ν gives a straight line with an intercept on the frequency axis at ν₀: below the threshold there is no emission at all, whatever the intensity. And the emission is instantaneous: it starts within about 10⁻⁹ s of the light falling on the surface, even at very low intensity.

NCERT sums up in four laws. (i) For a given metal and frequency above threshold, photocurrent ∝ intensity. (ii) For a given metal above threshold, K_max (or V₀) depends on frequency, not on intensity. (iii) Below the threshold frequency ν₀ there is no emission however intense the light; above it, K_max increases linearly with ν. (iv) Emission is instantaneous, with no detectable time lag.

In the wave picture, light is a continuous electromagnetic wave, its energy is spread uniformly over the wavefront, and a brighter (more intense) beam carries larger electric and magnetic field amplitudes. An electron at the surface absorbs energy continuously from the wave. Follow that picture through and it predicts three things, each of which contradicts what is seen. This three-point contrast is a standard 3-mark question, so learn it as a list.

1. Kinetic energy should rise with intensity. A more intense wave has larger fields and should push the electrons harder, so K_max should grow with intensity. Observed: K_max is independent of intensity; only the number of electrons (the current) changes.
2. There should be no threshold frequency. A wave of any frequency can deliver enough energy to free an electron if it is intense enough or shines for long enough. Observed: below ν₀ nothing comes out, however bright the light. And K_max depends on frequency, which has no role in the wave picture.
3. Emission should be delayed. Energy arriving spread over the wavefront is shared by all the electrons in the illuminated area, and the tiny share reaching any one electron takes time to build up to φ₀. NCERT's estimate is hours for a weak source. Observed: emission starts within 10⁻⁹ s even at very low intensity.

The failure is not in the details but in the assumption that energy is delivered continuously. The observations look as if the light energy is delivered in single lumps to single electrons, and each lump is bigger for higher frequency. That is exactly what Einstein proposed.

Einstein (1905) took Planck's idea one step further. Radiation of frequency ν is not a continuous wave but a stream of discrete energy packets, each of energy E = hν, where h = 6.63 × 10⁻³⁴ J s is Planck's constant. In photoelectric emission one quantum is absorbed by one electron, completely and at once. If the electron is at the surface, it spends φ₀ to get out and keeps the rest as kinetic energy. So the most energetic photoelectron has

K_max = hν − φ₀

This is Einstein's photoelectric equation. Electrons from deeper inside lose some energy in collisions on the way out and emerge with less than K_max, which is why there is a spread of kinetic energies up to this maximum. Since K_max = eV₀, the equation is also written eV₀ = hν − φ₀, or V₀ = (h/e)ν − φ₀/e.

Now check it against the four observations, in the order the marking scheme expects.

• Threshold frequency. K_max cannot be negative, so emission needs hν ≥ φ₀, i.e. ν ≥ ν₀ = φ₀/h. Below ν₀ no single photon has enough energy, and intensity cannot help because it only changes the number of photons. Equivalently, λ ≤ λ₀ = hc/φ₀.
• K_max depends linearly on ν and not on intensity. The equation contains ν and not the intensity. V₀ against ν is a straight line of slope h/e, the same for every metal, with intercept −φ₀/e on the V₀ axis and ν₀ on the ν axis.
• Photocurrent ∝ intensity. Intensity is the number of photons crossing unit area per second; more photons mean more one-to-one absorptions, more electrons, more current, but each electron still gets hν.
• Instantaneous emission. Absorption of a photon is a single, complete event; there is no build-up. Even a weak beam of photons above threshold emits electrons at once, just fewer of them.

Millikan tested the equation between 1906 and 1916 by measuring V₀ against ν for several alkali metals. The lines were straight, their slope gave h close to Planck's value, and this was taken as proof of the photon picture. Einstein received the 1921 Nobel Prize for this work. In a numerical, put everything in eV: hν in eV is 1240/λ(nm), subtract φ₀ in eV, and the result is both K_max in eV and V₀ in volts.

The photoelectric effect gives us the particle picture of light, and NCERT lists what a photon is. Learn these as four statements; they are asked directly.

1. In its interaction with matter, radiation behaves as if it is made of particles called photons.
2. Each photon has energy E = hν and momentum p = hν/c = h/λ, and travels with speed c. Its rest mass is zero.
3. All photons of light of a given frequency (or wavelength) have the same energy and momentum, whatever the intensity. Intensity of radiation in the photon picture is the number of photons crossing unit area per unit time; increasing intensity increases the number of photons per second, not the energy of each.
4. Photons are electrically neutral and are not deflected by electric or magnetic fields. In a photon–particle collision (such as a photon–electron collision) the total energy and total momentum are conserved, but the number of photons need not be conserved: a photon may be absorbed or a new one created.

This particle picture found further support in 1924, when A. H. Compton confirmed it by studying the scattering of X-rays by electrons, the Compton effect.

The energy of a visible photon is around 2 to 3 eV, which is why only a few volts of stopping potential are ever needed. A worked example the board likes (NCERT Example 11.1): monochromatic light of frequency 6.0 × 10¹⁴ Hz from a 2.0 × 10⁻³ W source. Energy per photon E = hν = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.98 × 10⁻¹⁹ J, about 2.49 eV. Photons emitted per second N = P/E = 2.0 × 10⁻³ / 3.98 × 10⁻¹⁹ = 5.0 × 10¹⁵ per second. That enormous number is why we never notice the graininess of ordinary light, just as we don't feel individual molecules in a wind.

Two more useful facts. Doubling the frequency doubles the photon energy and momentum; doubling the intensity doubles the photon count. And momentum p = E/c means a photon of wavelength λ carries momentum h/λ, the same relation de Broglie will turn round in the next section.

Radiation has a dual nature: it interferes and diffracts like a wave, yet in the photoelectric effect it behaves like particles. In 1924 Louis de Broglie argued from symmetry that nature should treat matter the same way: moving particles of matter should show wave-like properties under suitable conditions. His hypothesis is that a particle of momentum p has associated with it a wavelength

λ = h/p = h/(mv)

This is the de Broglie wavelength, and the wave is called a matter wave. The relation comes from the photon: for a photon E = hν and p = E/c = hν/c = h/λ, so λ = h/p; de Broglie proposed that the same relation holds for a material particle. Note the two consequences. λ is inversely proportional to momentum, so heavier or faster particles have shorter wavelengths, and λ does not depend on the charge or nature of the particle, only on p.

The derivation the board asks for is the wavelength of an electron accelerated from rest through a potential difference V. Its kinetic energy is K = eV, so p = √(2mK) = √(2meV) and

λ = h/√(2meV)

Substituting h = 6.63 × 10⁻³⁴ J s, m = 9.1 × 10⁻³¹ kg and e = 1.6 × 10⁻¹⁹ C gives λ = 1.227/√V nm, with V in volts. For V = 100 V, λ = 0.123 nm, which is the X-ray range and comparable to atomic spacings in a crystal, so electrons can be diffracted by crystals. The same steps give the general forms λ = h/√(2mK) for a particle of kinetic energy K, and λ = h/√(2mqV) for a charge q accelerated through V.

Why don't we see matter waves in daily life? NCERT Example 11.3: a 150 g ball at 30 m s⁻¹ has λ = 6.63 × 10⁻³⁴/(0.150 × 30) ≈ 1.5 × 10⁻³⁴ m, far too small compared with any opening or obstacle for diffraction to show. An electron at 5.4 × 10⁶ m s⁻¹ has λ = 6.63 × 10⁻³⁴/(9.1 × 10⁻³¹ × 5.4 × 10⁶) ≈ 0.135 nm, comparable to atomic spacing and to X-ray wavelengths. This is why the wave nature of an electron is significant and measurable, while a macroscopic object's is not.