JEE Mains Physics · Dual Nature of Radiation and Matter

Photoelectric Laws and Graphs

Frequency decides whether electrons leave and how fast the fastest ones move; intensity decides only how many leave; every photoelectric graph is read from eV₀ = hν − φ.

All Dual Nature of Radiation and Matter notes JEE Mains Physics strategy

Why this matters

Twenty-one PYQs, all multiple choice, and one from 2026. Fourteen test what the frequency and the intensity of the light each control, nine of them as assertion-reason or pick-the-true-statements questions. Seven read a graph: five a stopping-potential line against frequency, one a kinetic-energy line, and one photocurrent against voltage. There is almost no arithmetic; the marks go to knowing which quantity moves.

Concept 1 of 2: What frequency and intensity each control in the photoelectric effect

One photon frees at most one electron, and it does so at once. The energy of each photon is set by the frequency, so the frequency decides whether an electron can leave and how much energy the fastest one carries. The intensity only sets how many photons arrive, so it decides how many electrons leave, never how fast. A wave picture of light cannot explain any of this, which is why the effect is the evidence for photons.

Definition

One photon, one electron, no delay.
$K_{\max} = h\nu - \phi$ , and the stopping potential is $V_0 = K_{\max}/e$ .
$V_0$ depends on the frequency and on the metal. It does not depend on the intensity, the power of the source or its distance.
The photocurrent and the saturation current grow in proportion to the intensity, above threshold.
At a fixed intensity, a higher frequency means fewer photons per second, since $n = IA/h\nu$ .
To stop the electrons, the COLLECTOR is made negative with respect to the emitter.
$v_{\max}^{2}$ is linear in ν, but $v_{\max}$ itself is not.

Quantity	Raise the frequency (above threshold)	Raise the intensity (same frequency)
Maximum kinetic energy	Rises linearly: hν − φ	No change
Stopping potential	Rises linearly: (hν − φ)/e	No change Moving the lamp farther away dims it; the stopping potential stays the same.
Saturation current	Set by photons per second, not by their energy	Rises in proportion
Whether emission happens	Starts once ν passes ν₀	Never below ν₀, however bright
Delay before emission	None: emission is instant	None: emission is instant
Photons per second at fixed intensity	Falls, as n = IA/hν	Rises in proportion

Frequency sets the energy of each electron; intensity sets the number of electrons.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q101Moderate

Example 1 · Dual Nature of Radiation and Matter · Photoelectric Laws and Graphs

In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:

Brighter light, same stopping potential

Raising the intensity sends more photons of the same energy. More electrons leave, so the current rises, but the fastest of them is no faster, so the stopping potential is unchanged.

Doubling the frequency more than doubles the kinetic energy

K = hν − φ. At 2ν it becomes 2hν − φ = 2K + φ, which is more than 2K. The stopping potential likewise more than doubles.

The collector is made negative, not the emitter

Electrons are stopped by making the collecting plate negative with respect to the emitting surface. Making the emitter itself negative pushes electrons away from it and helps the current.

Drill 13 more on what frequency and intensity each control in the photoelectric effect

Concept 2 of 2: Reading photoelectric graphs

Rewrite Einstein's equation as V₀ = (h/e)ν − φ/e. That is a straight line in ν. Its slope is made only of constants of nature, so it is the same for every metal; the metal changes only where the line starts. The current-voltage curves follow from the laws: intensity raises the saturation current, frequency moves the cut-off voltage.

Definition

$eV_0 = h\nu - \phi$ , so $V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}$ .
The $V_0$ –ν line has slope h/e and meets the ν-axis at $\nu_0$ . Extended back, it meets the $V_0$ -axis at $-\phi/e$ .
The $K_{\max}$ –ν line has slope h.
Several metals on one plot give parallel lines. The line that starts at the lowest frequency has the smallest φ and gives the most energetic electrons for the same light.
To read the work function off a graph, find $\nu_0$ and use $\phi = h\nu_0$ , or read the $V_0$ -intercept.

Graph	Shape	Slope	Intercepts, and what shifts the graph
Stopping potential against frequency	Straight line from ν₀ upward	h/e, the same for every metal	Meets the ν-axis at ν₀ and, extended, the V₀-axis at −φ/e; a larger φ shifts it right, parallel
Maximum kinetic energy against frequency	Straight line from ν₀ upward	h, the same for every metal	Meets the ν-axis at ν₀ and, extended, the K-axis at −φ
Photocurrent against collector voltage, two intensities, one frequency	Rises, then flattens at a saturation current	Flat once saturated	Both cut off at the same −V₀; the brighter light saturates higher Same cut-off voltage means same frequency.
Photocurrent against collector voltage, two frequencies, one intensity	Rises, then flattens at a saturation current	Flat once saturated	The higher frequency cuts off at the more negative voltage; the saturation level is the same
Photocurrent against intensity	Straight line through the origin	Constant for one metal and one frequency	Stays at zero below threshold at any intensity
Stopping potential against intensity	Horizontal line	Zero	Its height is set by the frequency

Every line here comes from eV₀ = hν − φ.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q97Moderate

Example 2 · Dual Nature of Radiation and Matter · Photoelectric Laws and Graphs

The work functions of Aluminium and Gold are

4.1eV

and

5.1eV

respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

The slope is the same for every metal

The slope of the stopping-potential line is h/e, built only from constants of nature. A larger work function moves the line to the right but does not tilt it.

The potential-axis intercept is negative

Extended back to ν = 0, the stopping-potential line meets the potential axis at −φ/e, below the origin. Its size gives the work function in eV.

Lowest threshold, fastest electrons

For the same light, the metal whose line starts at the lowest frequency has the smallest work function, so it gives out the most energetic electrons.

Drill 6 more on reading photoelectric graphs

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Reference tables (2)

What frequency and intensity each control in the photoelectric effect6 rows

Quantity	Raise the frequency (above threshold)	Raise the intensity (same frequency)
Maximum kinetic energy	Rises linearly: hν − φ	No change
Stopping potential	Rises linearly: (hν − φ)/e	No change Moving the lamp farther away dims it; the stopping potential stays the same.
Saturation current	Set by photons per second, not by their energy	Rises in proportion
Whether emission happens	Starts once ν passes ν₀	Never below ν₀, however bright
Delay before emission	None: emission is instant	None: emission is instant
Photons per second at fixed intensity	Falls, as n = IA/hν	Rises in proportion

Frequency sets the energy of each electron; intensity sets the number of electrons.

Reading photoelectric graphs6 rows

Graph	Shape	Slope	Intercepts, and what shifts the graph
Stopping potential against frequency	Straight line from ν₀ upward	h/e, the same for every metal	Meets the ν-axis at ν₀ and, extended, the V₀-axis at −φ/e; a larger φ shifts it right, parallel
Maximum kinetic energy against frequency	Straight line from ν₀ upward	h, the same for every metal	Meets the ν-axis at ν₀ and, extended, the K-axis at −φ
Photocurrent against collector voltage, two intensities, one frequency	Rises, then flattens at a saturation current	Flat once saturated	Both cut off at the same −V₀; the brighter light saturates higher Same cut-off voltage means same frequency.
Photocurrent against collector voltage, two frequencies, one intensity	Rises, then flattens at a saturation current	Flat once saturated	The higher frequency cuts off at the more negative voltage; the saturation level is the same
Photocurrent against intensity	Straight line through the origin	Constant for one metal and one frequency	Stays at zero below threshold at any intensity
Stopping potential against intensity	Horizontal line	Zero	Its height is set by the frequency

Every line here comes from eV₀ = hν − φ.

Watch out for (6)

Brighter light, same stopping potential
→ What frequency and intensity each control in the photoelectric effect
Doubling the frequency more than doubles the kinetic energy
→ What frequency and intensity each control in the photoelectric effect
The collector is made negative, not the emitter
→ What frequency and intensity each control in the photoelectric effect
The slope is the same for every metal
→ Reading photoelectric graphs
The potential-axis intercept is negative
→ Reading photoelectric graphs
Lowest threshold, fastest electrons
→ Reading photoelectric graphs

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