In the wave picture of light, the intensity I of light is related to the amplitude A of the wave as : (a) Ipro

Question

I

(a)

I\propto\sqrt{A}

(b)

I\propto A

(c)

I\propto A^{2}

(d)

I\propto\dfrac{1}{A^{2}}

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Concept understanding — Photoelectric Effect

The Photoelectric Effect: When Light Knocks Electrons Loose

Imagine you're throwing tennis balls at a wall covered in loose pebbles. If you throw hard enough, a pebble might get knocked off. That's the basic picture — but the photoelectric effect is the quantum version of this, and it completely shattered classical physics.

The Intuition

Light is made of tiny packets of energy called photons. Each photon carries a specific amount of energy, determined by its colour (frequency). When a photon hits a metal surface, it can transfer its energy to an electron inside the metal. If that energy is enough, the electron breaks free and flies out.

Think of electrons in a metal like people in a room with a high window. To escape, they need enough energy to reach the window sill. A photon is like a boost — but only if it gives enough energy in one shot. No amount of weak boosts (dim light) will work if each individual boost is too small.

The Precise Statement

E_{\text{photon}} = hf = \phi + K_{\text{max}}

Where:

$E_{\text{photon}} = hf$ is the energy of a photon (Planck's constant $h = 6.63 \times 10^{-34} \ \text{J·s}$ , $f$ is frequency)
$\phi$ is the work function — the minimum energy needed to remove an electron from that metal
$K_{\text{max}}$ is the maximum kinetic energy of the ejected electron

What Classical Physics Got Wrong

Before Einstein (1905), physicists thought light was a continuous wave. They expected:

Brighter light → more energy per electron → faster electrons
Any colour would eventually eject electrons if you waited long enough

But experiments showed the opposite:

Observation	Classical Prediction	Actual Result
Effect of intensity	Brighter light → faster electrons	Brighter light → more electrons, same speed
Threshold frequency	None — any light works eventually	Below a certain frequency, no electrons no matter how bright
Time delay	Electrons need time to absorb energy	Electrons appear instantly (within $10^{-9}$ s)

The Key Insight

Einstein said: light behaves like a stream of particles (photons), each with energy $hf$ . One photon interacts with one electron. If $hf < \phi$ , the electron cannot escape — period. If $hf > \phi$ , the excess energy becomes kinetic energy:

K_{\text{max}} = hf - \phi

This is why:

Increasing intensity (more photons) ejects more electrons, but each electron still gets the same energy per photon — so their speed doesn't change.
Below threshold frequency, even a trillion photons per second can't help — each one is too weak individually.

Important

The photoelectric effect proved that light is quantized — it comes in discrete packets. This was the birth of quantum mechanics. Einstein won the 1921 Nobel Prize for this, not for relativity.

A Worked Example

Note

Problem: A metal has work function $\phi = 2.0 \ \text{eV}$ . Light of frequency $f = 6.0 \times 10^{14} \ \text{Hz}$ shines on it. Find the maximum kinetic energy of ejected electrons. ( $h = 4.14 \times 10^{-15} \ \text{eV·s}$ )

Step 1: Photon energy

$E = hf = (4.14 \times 10^{-15})(6.0 \times 10^{14}) = 2.48 \ \text{eV}$ …

Why this formula?

Photoelectric Effect: Why the Key Formulas Hold

The photoelectric effect is a cornerstone of quantum physics. It showed that light behaves as particles (photons) , not just waves. Let's build the reasoning step-by-step.

1. The Core Idea: Energy Conservation

When a photon hits a metal surface, it transfers all its energy to a single electron inside the metal.

The photon's energy is $E = hf$ , where $h$ is Planck's constant and $f$ is the frequency of light.
The electron needs a minimum energy to escape the metal — this is called the work function, $\phi$ .

Why only one electron?

Einstein proposed that light is quantized into discrete packets (photons). A single photon cannot split its energy among multiple electrons — it interacts with one electron at a time.

2. The Photoelectric Equation

If the photon's energy is greater than the work function, the excess energy becomes the electron's kinetic energy after escape:

hf = \phi + K_{\text{max}}

Where:

$hf$ = energy of incident photon
$\phi$ = work function (minimum energy to remove electron)
$K_{\text{max}}$ = maximum kinetic energy of ejected electron

Why "maximum" kinetic energy?

Electrons inside the metal have different binding energies.
Some electrons are near the surface (loosely bound) → get maximum $K$ .
Others are deeper → lose energy in collisions before escaping → lower $K$ .

3. The Stopping Potential Connection

We measure $K_{\text{max}}$ using a stopping potential $V_s$ :

K_{\text{max}} = eV_s

Where $e$ is the electron charge. This is because:

An electric field opposing the electron's motion does work $eV_s$ to stop it.
At the stopping potential, the electron's kinetic energy is exactly balanced by the electric potential energy.

Combining:

hf = \phi + eV_s

This is the Einstein photoelectric equation in its most testable form.

4. Why the Threshold Frequency Exists

From the equation:

hf = \phi + eV_s

If $f$ is too low, $hf < \phi$ . Then:

The photon cannot supply enough energy to overcome the work function.
No electron is ejected, regardless of light intensity.

The threshold frequency $f_0$ is when $K_{\text{max}} = 0$ :

hf_0 = \phi \quad \Rightarrow \quad f_0 = \frac{\phi}{h}

Why intensity doesn't matter for ejection?

Intensity = number of photons per second.
Each photon still has energy $hf$ . If $hf < \phi$ , even a billion photons won't eject an electron — each photon is individually too weak.

5. Why Kinetic Energy Depends on Frequency, Not Intensity

From $K_{\text{max}} = hf - \phi$ :

Frequency $f$ directly determines $K_{\text{max}}$ .
Intensity only affects the number of electrons ejected (more photons → more electrons), not their individual energy.

This was the key experimental contradiction with classical wave theory:

Classical: Higher intensity = bigger wave amplitude = more energy to electrons.
Reality: Higher frequency = more energy per electron; intensity only changes current.

6. Summary of Key Relationships …

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In the wave picture of light, intensity is proportional to the square of the amplitude: $I \propto A^2$ . This follows from the fact that intensity is the energy transported per unit area per unit time, and the energy of a wave is proportional to the square of its amplitude.

The key idea here is that light, when treated as a wave, carries energy. The intensity of light is defined as the power (energy per second) passing through a unit area perpendicular to the direction of propagation. For any mechanical or electromagnetic wave, the energy carried is directly linked to how "big" the wave is — that is, its amplitude.

Think of a wave on a string: if you pull the string up higher (larger amplitude), it takes more energy to create that wave. In fact, the energy stored in a wave is proportional to the square of its amplitude. The same principle applies to light waves, which are electromagnetic waves. The electric and magnetic fields oscillate, and the energy density of the field is proportional to the square of the field strength (amplitude). So, intensity, being the flow of that energy, also ends up proportional to $A^2$ .

Let’s walk through the reasoning step by step.

Definition of intensity for a wave

Intensity $I$ is defined as the average power transferred per unit area. For a wave, this is the rate at which energy flows through a surface. Mathematically, for a plane wave, $I = \frac{\text{Power}}{\text{Area}}$ .
Energy density of an electromagnetic wave

In the wave picture, light consists of oscillating electric and magnetic fields. The energy per unit volume (energy density) stored in an electromagnetic wave is given by $u = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2}\frac{B^2}{\mu_0}$ . For a wave, $E$ and $B$ are related by $B = E/c$ , so the total energy density simplifies to $u = \epsilon_0 E^2$ (averaged over time). Here, $E$ is the amplitude of the electric field — which is the amplitude $A$ of the wave.
Relating energy density to intensity

The intensity is the energy density multiplied by the speed of the wave: $I = u \cdot c$ . Since $u \propto E^2$ , we get $I \propto E^2$ . The amplitude $A$ of the wave is exactly this electric field amplitude $E$ (or proportional to it). Therefore, $I \propto A^2$ .
Why not the other options? …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 113 on this concept.

CBSE 2026Set 55/2/11 markMCQ
$\phi_0$
›Reveal solutionSolution
The maximum kinetic energy of photoelectrons is determined by Einstein's photoelectric equation: it depends on both the photon energy (set by frequency $\nu$ ) and the work function $\phi_0$ of the material. The answer is (D).

Why the photoelectric effect reveals energy quantization

When light strikes a metal surface, electrons can be ejected—but not in the way classical wave theory predicted. Einstein's revolutionary insight was that light arrives in discrete packets (photons), each carrying energy $E = h\nu$ . An electron absorbs one photon entirely; if that energy exceeds the minimum needed to escape the metal (the work function $\phi_0$ ), the electron breaks free, and any leftover energy becomes kinetic energy.

This is fundamentally an energy-balance problem. The photon delivers a fixed amount of energy $h\nu$ . The electron must "pay" $\phi_0$ to escape. What remains is the maximum kinetic energy:
$K_{\text{max}} = h\nu - \phi_0$
Notice what this equation tells us: $K_{\text{max}}$ increases linearly with frequency $\nu$ and decreases with larger work function $\phi_0$ . Both parameters matter.

Step-by-step reasoning

The photon energy is $h\nu$ .

A single photon of frequency $\nu$ carries energy proportional to that frequency. Higher frequency means more energetic photons (ultraviolet photons pack more punch than red photons).

The work function $\phi_0$ is the escape barrier.

Different materials bind their electrons with different strengths. Cesium has a low work function (~2 eV), so even visible light can eject electrons. Platinum has a high work function (~6 eV), requiring ultraviolet light. This material property directly subtracts from the available kinetic energy.

Energy conservation gives $K_{\text{max}} = h\nu - \phi_0$ .

The electron that escapes with maximum kinetic energy is one that was initially at the Fermi level (loosest bound) and lost no energy to collisions on the way out. All the "profit" after paying $\phi_0$ goes into kinetic energy.

Intensity does not affect $K_{\text{max}}$ .

Intensity measures the number of photons arriving per unit time, not the energy of each photon. Brighter light ejects more electrons (higher photocurrent), but each electron still gets energy from just one photon. Doubling intensity doubles the electron count, not their individual speeds. …
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CBSE 2026Set 55/2/11 markMCQ
$\nu\,(>\nu_0)$
›Reveal solutionSolution
The Assertion is true because photocurrent depends on the number of photoelectrons, which increases with light intensity. The Reason is also true — stopping potential depends on frequency, not intensity. But the Reason does not explain the Assertion, since they involve different physical mechanisms. So the correct choice is (B).

The core physics: photoelectric effect and the two separate ideas

The photoelectric effect has a clean conceptual split that many students miss. There are two independent things happening when light hits a metal surface:

How many electrons come out — this is governed by the intensity of light (number of photons per second). More photons → more electrons ejected → larger photocurrent.

How much kinetic energy each electron has — this is governed by the frequency of light (energy per photon). Higher frequency → more energy per electron → you need a larger stopping potential to bring them to rest.

These two are completely separate. Intensity does not affect the energy of individual electrons; frequency does not affect how many electrons are ejected (above threshold). This is the key insight that makes the question straightforward.

Step-by-step reasoning

1. Understanding the Assertion (A)

On increasing the intensity of incident light of frequency $\nu > \nu_0$ on a photosensitive surface, the photocurrent increases.

When frequency is above the threshold $\nu_0$ , every photon that hits the surface can eject a photoelectron (assuming it's absorbed). Intensity means the number of photons per second per unit area. If you increase intensity, more photons arrive each second, so more electrons are knocked out. The photocurrent (rate of flow of charge) is directly proportional to the number of electrons emitted per second.
$\propto$
So the Assertion is true.

2. Understanding the Reason (R)

The stopping potential for a photosensitive surface increases with increase of frequency $\nu > \nu_0$ of incident light.

Stopping potential $V_s$ is the voltage needed to stop the most energetic photoelectrons. Einstein's photoelectric equation gives:
$K_{\text{max}} = h\nu - \phi_0$
where $\phi_0$ is the work function. Since $K_{\text{max}} = eV_s$ , we have:
$eV_s = h\nu - \phi_0 \quad \Rightarrow \quad V_s = \frac{h}{e}\nu - \frac{\phi_0}{e}$
This is a straight line with slope $h/e$ . As $\nu$ increases, $V_s$ increases linearly. So the Reason is also true. …
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CBSE 2026Set 55/3/11 markMCQ
$200$
›Reveal solutionSolution
The fastest photoelectron’s kinetic energy is found from Einstein’s photoelectric equation: $K_{\text{max}} = h\nu - \phi$ . Converting wavelength to frequency and using the given work function gives $K_{\text{max}} \approx 2.0$ eV, so the correct option is (D).

The photoelectric effect is a clean, direct application of energy conservation: a photon gives all its energy to an electron. The electron uses some of that energy to escape the surface (the work function $\phi$ ), and whatever remains shows up as kinetic energy. The fastest electron is the one that loses the least energy on its way out — so its kinetic energy is simply $h\nu - \phi$ .

Let’s walk through it.

Find the photon energy in eV. Wavelength $\lambda = 200$ nm $= 200 \times 10^{-9}$ m. Photon energy $E = h\nu = \frac{hc}{\lambda}$ . Use the handy constant $hc = 1240$ eV·nm (this is a standard shortcut for such problems). So

$E = \frac{1240\ \text{eV·nm}}{200\ \text{nm}} = 6.2\ \text{eV}.$
Tip

Memorise $hc \approx 1240$ eV·nm — it converts wavelength in nm directly to energy in eV without messing with SI units.

Apply Einstein’s photoelectric equation. The maximum kinetic energy of emitted photoelectrons is

$K_{\text{max}} = h\nu - \phi.$
Here $\phi = 4.2$ eV. So

$K_{\text{max}} = 6.2\ \text{eV} - 4.2\ \text{eV} = 2.0\ \text{eV}.$ …
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CBSE 2026Set 55/3/11 markMCQ
$Q. Assertion (A) : Photoelectric current depends upon the intensity of the incident radiation. Reason (R) : Stopping potential is independent of the intensity of the incident radiation. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.$
›Reveal solutionSolution
The photoelectric current depends on intensity (more photons → more electrons), while stopping potential depends only on frequency (photon energy minus work function). Both statements are true, but the reason does not explain the assertion — they are independent facts about the photoelectric effect. The correct option is (B).

The photoelectric effect is one of those rare topics where a single experiment cleanly separates two ideas that students often mix up: how many electrons get ejected versus how energetic each electron is. The assertion and reason here test exactly that distinction.

The core physics: When light hits a metal surface, each photon can knock out one electron — but only if the photon’s energy ( $hf$ ) exceeds the metal’s work function ( $\phi$ ). The number of electrons ejected (and hence the current) depends on how many photons arrive per second — that’s intensity. But the maximum kinetic energy of each ejected electron depends only on the photon’s frequency, not on how many photons there are.

Let’s walk through each statement carefully.

Assertion (A): Photoelectric current depends on the intensity of incident radiation.

Current is charge per second. Each ejected electron carries charge $e$ . So if more electrons leave the metal per second, the current increases. What controls the number of electrons per second? The number of photons striking the metal per second — that is, the intensity (for a fixed frequency).

Tip

Think of it like rain: intensity is how hard it’s raining (number of raindrops per second). More raindrops → more splashes (electrons). But each splash’s height depends on the size of the raindrop (frequency), not on how many are falling.

So (A) is true.

Reason (R): Stopping potential is independent of the intensity of incident radiation.

Stopping potential $V_0$ is the voltage needed to stop the most energetic photoelectrons. Einstein’s photoelectric equation gives:

$eV_0 = hf - \phi$
Here $f$ is the frequency of light, $\phi$ is the work function. Notice: intensity does not appear anywhere in this equation. Whether you shine a dim light or a bright light of the same colour, the stopping potential stays the same. …
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CBSE 2026Set DS1 mark
$Q. The work function of a metal is 3.3 eV. Calculate the minimum frequency of photon that will emit the photoelectron.$
›Reveal solutionSolution

The minimum (threshold) frequency is $\nu_0 = W/h \approx 8.0\times10^{14}$ Hz.

Concept. Photoelectrons are emitted only when the incident photon's energy at least equals the work function $W$ of the metal. The minimum (threshold) frequency corresponds to a photon whose whole energy is used up just to free the electron, with no kinetic energy left:

$h u_0 = W \;\Rightarrow\; u_0 = \frac{W}{h}.$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. When light falls on a metal surface, the maximum kinetic energy of the emitted electrons depends upon$
(a) the time for which light falls on the metal
(b) the frequency of the incident light
(c) the intensity of the incident light
(d) the velocity of the incident light
›Reveal solutionSolution

Maximum kinetic energy of photoelectrons depends only on the frequency of incident light (and the metal's work function), never on intensity or time.

By Einstein's photoelectric equation,

$KE_{max} = h\nu - \phi_0$

where $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\phi_0$ is the work function of the metal (a fixed property of the metal surface). Increasing the intensity of light (at fixed frequency) increases the number of photoelectrons emitted per second (photocurre …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The work function of four metals P, Q, R and S are 2.3 eV, 3.2 eV, 4.25 eV and 5.15 eV respectively. Among these the metal having lowest threshold frequency will be$
(a) R
(b) S
(c) Q
(d) P
›Reveal solutionSolution

Threshold frequency is directly proportional to work function (nu0 = W/h), so the metal with the smallest work function has the lowest threshold frequency.

Einstein's photoelectric equation gives the threshold frequency as nu0 = W/h, where W is the work function and h is Planck's constant (a fixed constant). Since nu0 scales directly with W, the metal with the smallest work function has the smallest ( …

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CBSE 2026Set ANNUAL1 mark
$Q. If the workfunction of a metal is 4.5 eV and the kinetic energy of the photoelectron emitted from its surface is 1.5 eV, then calculate the energy of a photon of the incident light.$
›Reveal solutionSolution

Einstein's photoelectric equation says the photon's energy is used partly to free the electron (work function) and partly becomes its kinetic energy; adding the two gives the photon energy.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Photoelectric effect supports$
(a) wave nature of light
(b) particle nature of light
(c) dual nature of light
(d) polarization nature of light
›Reveal solutionSolution

Classical wave theory cannot explain the instantaneous emission or the frequency threshold seen in the photoelectric effect; only Einstein's photon (particle) picture of light does.

In the photoelectric effect, light falling on a metal surface ejects electrons. Key observations - that emission starts INSTANTLY (no time lag) even at very low intensity, that there is a definite THRESHOLD FREQUENCY below which no electrons are emitted no matter how intense the light, and that the maximum kinetic energy of emitted electrons depends on FREQUENCY (not intensity) - cannot be explained by the classical wave theory of light. Einstein explained these features by treating light as made of discrete quanta (photons) of …

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CBSE 2026Set ANNUAL1 mark
$Q. Define work function of metal.$
›Reveal solutionSolution

Work function is the minimum energy needed to pull the least tightly bound electron out of a metal surface.

Electrons inside a metal are held back from escaping by attractive forces from the positive ions in the lattice. The work function phi_0 is defined as the minimum energy that must be supplied to an electron (typically the most energetic, least tightly bound one, at the Fermi level) so that it can just escape from the metal surface, with no kinetic energy left over. It is usually expressed in electron-volts (eV) and is related to the thre …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Stopping potential is minimum for:$
(a) Yellow
(b) Blue
(c) Violet
(d) Red
›Reveal solutionSolution

Stopping potential increases with the frequency of incident light; red light has the lowest frequency among the options, so it gives the minimum stopping potential.

By Einstein's photoelectric equation, $eV_0 = h\nu - \phi_0$ , so stopping potential $V_0$ increases linearly with frequency $\nu$ . Among yellow, blue, violet and red light, red has the longest wavel …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The work function of caesium metal is 2.14 eV. When light of frequency 6 \times 10^14 Hz is incident on the metal surface, photoemission of electrons occurs. What is the Stopping potential?$
(a) 34 V
(b) 3.4 V
(c) 340 V
(d) 0.34 V
›Reveal solutionSolution

Using Einstein's photoelectric equation, the stopping potential comes out to about 0.34 V.

Step 1 — photon energy:

$E = h\nu = (6.626\times10^{-34})(6\times10^{14}) = 3.976\times10^{-19}\ \text{J}$

Converting to eV: $\dfrac{3.976\times10^{-19}}{1.602\times10^{-19}} \approx 2.48\ \text{eV}$

Step 2 — Einstein's photoelectric equation:

$K_{max} = h\nu - \phi_0 = 2.48 - 2.14 = 0.34\ \text{eV}$

…

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Q.In the wave picture of light, the intensity III of light is related to the amplitude AAA of the wave as :

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Q.In the wave picture of light, the intensity $I$ of light is related to the amplitude $A$ of the wave as :

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