Photoelectric Effect and Wave Theory of Light

Dual Nature of Radiation and Matter · CBSE

Theory

What this section has to deliver

This is CBSE's standard 5-mark question: "Why is the wave theory of electromagnetic radiation not able to explain the photoelectric effect? How does the photon picture resolve this problem?" It also appears as a 2- or 3-mark question asking for two or three features that wave theory cannot explain. Structure the answer as: what the wave picture says, three specific contradictions, then the photon resolution.

What the wave picture of light says

By the end of the nineteenth century the wave nature of light was firmly established: interference, diffraction and polarisation are all explained naturally by it.
In this picture light is an electromagnetic wave whose electric and magnetic fields carry energy distributed continuously over the whole region of space through which the wave extends.
Greater intensity means greater amplitude of the electric and magnetic fields, hence more energy delivered per unit area per unit time.
Absorption takes place continuously over the entire wavefront, and every electron on the illuminated surface shares in the arriving energy.

Contradiction 1: dependence of $K_{max}$ on intensity

Wave theory predicts: the more intense the radiation, the greater the amplitude of the fields, so the greater the energy absorbed by each electron, so the greater the maximum kinetic energy of the photoelectrons. $K_{max}$ should increase with intensity.
Experiment shows: $K_{max}$ , measured through the stopping potential, is completely independent of intensity. Doubling the intensity leaves $V_0$ exactly where it was and only raises the saturation current.

Contradiction 2: the existence of a threshold frequency

Wave theory predicts: light of any frequency whatever should eject electrons, provided the beam is intense enough, or is left on long enough, to supply the escape energy. No threshold frequency should exist.
Experiment shows: below the threshold frequency $\nu_0$ of the metal, no electron is ever emitted, no matter how intense the light or how long it shines.

Contradiction 3: the time lag

Wave theory predicts: the energy of the wave is spread continuously and uniformly over the wavefront, and a very large number of electrons share it, so the energy absorbed per electron per unit time is tiny. Explicit calculations give hours or more for one electron to accumulate enough energy to overcome the work function. There should be a clear, intensity-dependent delay before emission begins.
Experiment shows: emission is instantaneous. The time lag is of the order of $10^{-9}$ s or less, even when the incident radiation is made exceedingly dim.

How the photon picture resolves all three

Radiation of frequency $\nu$ is not a continuous flow of energy but a stream of discrete packets — photons — each carrying exactly $h\nu$ .
In the elementary process, one electron absorbs one whole photon. It is all-or-nothing: the electron takes $h\nu$ or nothing, and it cannot collect fractions of energy from many photons.
Hence $K_{max} = h\nu - \phi_0$ : the maximum kinetic energy is fixed by the energy of one photon and the work function alone, and cannot depend on how many photons arrive. Intensity has no way into this equation. Contradiction 1 removed.
Hence emission requires $h\nu > \phi_0$ , i.e. $\nu > \nu_0 = \phi_0/h$ . If a single photon is too feeble, sending more of them cannot help, because each electron still meets only one photon at a time. Contradiction 2 removed.
Hence emission is instantaneous: the absorption of a single quantum by a single electron is a one-step process with nothing to accumulate. Low intensity means fewer electrons taking part, not a longer wait. Contradiction 3 removed.
Increasing the intensity increases the number of photons per unit area per unit time, so more electrons are liberated per second and the photocurrent rises in proportion — which is exactly the observed intensity dependence of the current.

The two-line summary if you are short of time

Observation	Wave theory	Photon picture
$K_{max}$ independent of intensity	fails: predicts increase with intensity	$K_{max}=h\nu-\phi_0$ , no intensity in it
Threshold frequency exists	fails: predicts emission at any frequency	needs $h\nu>\phi_0$ for one photon
Emission instantaneous	fails: predicts a delay of hours	one photon absorbed whole, in one step
Current proportional to intensity	partly consistent	more photons per second, more electrons per second

NCERT's own remark, worth a line: the observations on the stopping potential — its independence of intensity and its dependence on frequency — are the crucial discriminator between the wave picture and the photon picture.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Why is wave theory of electromagnetic radiation not able to explain the photoelectric effect? How does the photon picture resolve this problem? OR Discuss briefly how the wave theory of light cannot explain the photoelectric effect. OR Write three observed features of photoelectric effect which cannot be explained by the wave theory of light.
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2019 / 2/3, AI 2015 / 2/5, AI 2015C, An · Section E, 5 marksAnswer 1 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
The observed characteristics of the photoelectric effect could not be explained on the basis of the wave theory of light: (i) According to wave theory, light propagates as wavefronts and its energy is distributed uniformly over the wavefront. As the intensity of light increases, the amplitude of the waves (and hence the energy stored in them) increases, so more energy would be delivered to the electrons of the metal, and consequently the kinetic energy of the emitted electrons should increase with intensity. But experimentally, the maximum kinetic energy of photoelectrons is found to be INDEPENDENT of the intensity of incident light. (ii) According to wave theory, light of ANY frequency should be able to eject electrons, provided the intensity is high enough to supply the energy needed for emission. But experimentally, light of frequency below the threshold frequency cannot eject electrons no matter how large the intensity. (iii) According to wave theory, since energy is spread continuously and uniformly over the wavefront (not concentrated on particular electrons), electrons would need to absorb energy gradually over some finite time before accumulating enough to escape — so there should be an observable time lag between incidence of light and emission of electrons, more so for weaker intensity. Experimentally, however, photoelectric emission is an instantaneous process (time lag $<10^{-9}$ s), even for very low intensities. The photon picture resolves this: light of frequency $\nu$ consists of discrete packets (photons) of energy $h\nu$ each. In the interaction of a photon with a free electron, the ENTIRE energy of one photon is absorbed by one electron (all-or-nothing), instantaneously. If this quantum $h\nu$ exceeds the work function $W_0$ (the minimum energy needed to escape), the electron is emitted immediately with excess kinetic energy $K_{max}=h\nu-W_0$ — a photon of frequency below threshold simply cannot free an electron regardless of how many photons (intensity) arrive, since each electron interacts with only one photon at a time. Increasing intensity increases the NUMBER of photons (and hence photoelectrons) per second, not the energy per photon — hence $K_{max}$ is independent of intensity, exactly as observed.