Q.Consider a beam of electrons (each electron with energy E0) incident on a metal surface kept in an evacuated chamber. Then
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
Ephoton=hf=ϕ+Kmax
Where:
- Ephoton=hf is the energy of a photon (Planck's constant h=6.63×10−34 J⋅s, f is frequency)
- ϕ is the work function — the minimum energy needed to remove an electron from that metal
- Kmax 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<ϕ, the electron cannot escape — period. If hf>ϕ, the excess energy becomes kinetic energy:
Kmax=hf−ϕ
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.
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
Problem: A metal has work function ϕ=2.0 eV. Light of frequency f=6.0×1014 Hz shines on it. Find the maximum kinetic energy of ejected electrons. (h=4.14×10−15 eV⋅s)
Step 1: Photon energy
E=hf=(4.14×10−15)(6.0×1014)=2.48 eV
Step 2: Subtract work function
Kmax=2.48−2.0=0.48 eV
Step 3: Convert to joules if needed
0.48 eV×1.6×10−19=7.68×10−20 J
The electron escapes with this much kinetic energy.
Common Mistake to Avoid
Students often think "more intense light means more energy per electron." Wrong. Intensity = number of photons per second. Each photon still has the same hf. More photons = more electrons, but each electron gets the same energy kick.
The Big Picture
The photoelectric effect is your first encounter with wave-particle duality. Light, which we model as a wave for interference and diffraction, behaves as a particle when transferring energy to matter. This duality is central to all of quantum mechanics.
Final takeaway: Light ejects electrons only if each photon carries enough energy individually. The colour (frequency) determines whether ejection happens; the brightness (intensity) determines how many electrons get ejected.
"Photoelectric effect formula and Einstein equation" is among the most-searched Class 12 physics topics, and it is a core result of the Dual Nature of Radiation and Matter chapter in the NCERT/CBSE Class 12 Physics curriculum. Work function and threshold frequency questions built on this concept appear in nearly every JEE Main and NEET physics paper.
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, ϕ.
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=ϕ+Kmax
Where:
- hf = energy of incident photon
- ϕ = work function (minimum energy to remove electron)
- Kmax = 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 Kmax using a stopping potential Vs:
Kmax=eVs
Where e is the electron charge. This is because:
- An electric field opposing the electron's motion does work eVs to stop it.
- At the stopping potential, the electron's kinetic energy is exactly balanced by the electric potential energy.
Combining:
hf=ϕ+eVs
This is the Einstein photoelectric equation in its most testable form.
4. Why the Threshold Frequency Exists
From the equation:
hf=ϕ+eVs
If f is too low, hf<ϕ. Then:
- The photon cannot supply enough energy to overcome the work function.
- No electron is ejected, regardless of light intensity.
The threshold frequency f0 is when Kmax=0:
hf0=ϕ⇒f0=hϕ
Why intensity doesn't matter for ejection?
- Intensity = number of photons per second.
- Each photon still has energy hf. If hf<ϕ, 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 Kmax=hf−ϕ:
- Frequency f directly determines Kmax.
- 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
| Quantity | Formula | Why it holds |
|---|---|---|
| Photon energy | E=hf | Light is quantized (Planck-Einstein) |
| Work function | ϕ=hf0 | Minimum energy to escape at threshold |
| Max kinetic energy | Kmax=hf−ϕ | Energy conservation per photon-electron |
| Stopping potential | eVs=hf−ϕ | Electric work balances kinetic energy |
| Threshold frequency | f0=ϕ/h | Below this, no ejection possible |
7. The Deeper "Why" — Particle Nature of Light
The photoelectric effect cannot be explained by classical wave theory because:
- Waves spread energy over the whole wavefront — an electron would take time to absorb enough energy.
- But experiments show instantaneous ejection (within 10−9 s).
- Wave theory predicts kinetic energy should increase with intensity — it doesn't.
Einstein's photon model resolves all three:
- Instantaneous — one photon, one interaction.
- Frequency-dependent — photon energy is hf.
- Intensity-independent — more photons = more electrons, not more energy per electron.
Key takeaway: The photoelectric effect is a direct consequence of energy quantization — both light and electron binding energy are quantized. The formulas are simply conservation laws applied to this quantum world.
Concept: Photoelectric effect vs. electron-impact (secondary) emission.
A photon is not the only particle that can eject a bound electron. An incident electron of kinetic energy E0 can also knock electrons out of the metal by collision (secondary emission). In such a collision the incident electron may hand over any fraction of its energy, from almost none up to nearly all of E0. The freed electron must still spend the work function ϕ to leave the surface, so the largest kinetic energy it can carry away is E0−ϕ.
Hence electrons are emitted with a range of energies, up to a maximum of E0−ϕ.
The correct option is (C): electrons can be emitted with any energy, with a maximum of E0−ϕ.
An incident electron beam ejects electrons by collision (secondary emission); the freed electrons come out with a spread of energies whose maximum is E0−ϕ — option (C).
Setting up the physics
The photoelectric effect is the ejection of electrons by photons, but this question is different: the metal is bombarded by a beam of electrons, each of energy E0. Electrons are charged particles and interact with the metal's electrons through the Coulomb force, so an incident electron can transfer energy to a bound electron in a collision and knock it out. This is secondary electron emission — a real, well-known process (it is exactly what multiplies the signal on the dynodes of a photomultiplier).
How much energy can an ejected electron have?
- Energy available. The incident electron brings kinetic energy E0.
- Energy that must be paid. To escape the metal, the freed electron must spend at least the work function ϕ.
- The transfer is not fixed. In a collision the incident electron can give up any fraction of its energy — a glancing hit transfers little, a head-on hit transfers a lot. So the freed electron can emerge with kinetic energy anywhere from 0 up to a maximum.
- The maximum. The most the ejected electron can retain is the incident energy minus the escape cost:
Kmax=E0−ϕ.
So the emitted electrons are not mono-energetic; they form a continuous distribution up to E0−ϕ.
Why the other options fail
- (A) "No electrons emitted" is wrong: an energetic electron beam does eject electrons by collision.
- (B) "All with energy E0" is wrong: the incident electron loses part of its energy in the collision, and the escaping electron also pays ϕ.
- (D) "Maximum of E0" ignores the work function that must be spent to leave the surface.
The correct option is (C): electrons can be emitted with any energy, with a maximum of E0−ϕ.
Method: Distinguishing Emission Mechanisms — Photon Absorption vs Particle-Impact Collision
Use this reasoning pattern whenever a question describes electrons being knocked out of a metal by something other than a beam of light, and asks you to compare it with the ordinary photoelectric effect.
Steps
Step 1: Check what is actually incident on the metal
The photoelectric equation Ephoton=hf=ϕ+Kmax applies strictly to photon absorption — one photon, one electron, all-or-nothing. If the question instead describes a beam of charged particles (electrons, in this case) hitting the surface, that equation does not apply as written; you're dealing with a collision process instead.
Step 2: Recognise that a particle collision can transfer any fraction of energy
Unlike a photon (which either transfers all its energy or none), an incident particle interacting via a real physical collision can hand over anywhere from a small fraction to nearly all of its kinetic energy, depending on the geometry of the collision (glancing vs head-on). This means the freed electrons will not be mono-energetic — they emerge with a spread of energies.
Step 3: Find the maximum by subtracting the fixed escape cost once
Whatever the mechanism of energy transfer, an electron still has to pay the same fixed price — the work function ϕ — to leave the metal surface. So the maximum possible kinetic energy of an emitted electron is always
Kmax=(energy available to the electron)−ϕ
Here the energy available is the full incident energy E0 of the electron doing the knocking, so Kmax=E0−ϕ.
Step 4: Rule out options that violate either principle
Check every option against the two principles above: an option claiming "no emission" ignores that particle collisions do transfer energy; an option claiming a single fixed energy ignores that collisions transfer a variable amount; an option that forgets to subtract ϕ ignores that escape always has a cost.
Showing the 12 most recent of 21 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.The work function of a material is 6.6 eV. Then, the threshold wavelength of the metal is approximately (Take h=6.6×10−34 J.s) (A) 108 nm (B) 188 nm (C) 208 nm (D) 228 nm (E) 250 nm
›Reveal solutionSolution
Threshold wavelength satisfies W=hc/λ0. With W=6.6 eV this gives about 188 nm.
Convert the work function to joules:
W=6.6 eV=6.6×1.6×10−19=1.056×10−18 J.
Threshold wavelength:
λ0=Whc=1.056×10−18(6.6×10−34)(3×108)=1.056×10−181.98×10−25≈1.875×10−7 m.
That is ≈188 nm.
✓Final answerThe correct option is (B).
- KEAM 2026Set eng-2026-04184 marksMCQQ.Two light rays of wavelength λ and 4λ incident on the surface of a photo sensitive material emit electrons with max kinetic energy E and 6E respectively. The work function of the material is (h = Planck's constant, c = velocity of light in free space) (A) λhc (B) 2λhc (C) 5λ2hc (D) 5λ3hc (E) 3λhc
›Reveal solutionSolution
Writing Einstein's equation for both wavelengths and eliminating E gives the work function W=5λ2hc.
Einstein's photoelectric equation, KE=λhc−W:
For wavelength λ:
E=λhc−W(1)
For wavelength λ/4 (photon energy 4hc/λ):
6E=λ4hc−W(2)
Subtract (1) from (2):
5E=λ3hc ⇒ E=5λ3hc
From (1):
W=λhc−E=λhc−5λ3hc=5λ2hc
✓Final answerThe correct option is (C).
- KEAM 2026Set eng-2026-04194 marksMCQQ.If the threshold wavelengths of two metals are in the ratio 1:3, then the work functions of these metals are in the ratio (A) 1:3 (B) 2:1 (C) 3:1 (D) 1:2 (E) 3:2
›Reveal solutionSolution
Work function is inversely proportional to threshold wavelength, so wavelengths 1:3 give work functions 3:1.
The threshold (work function) relation is
ϕ=λ0hc⇒ϕ∝λ01.
Given λ0,1:λ0,2=1:3,
ϕ2ϕ1=λ0,1λ0,2=13=3:1.
✓Final answerThe correct option is (C).
- KEAM 2026Set eng-2026-04204 marksMCQQ.Stopping potentials for the metals A and B are 0.4 V and 1.6 V, respectively. When illuminated by same light, the difference in their work functions is: (A) 2.0 eV (B) 1.2 eV (C) 6.4 eV (D) 0.4 eV (E) 2.0 V
›Reveal solutionSolution
With identical incident light, the work-function difference equals e times the stopping-potential difference: 1.2 eV.
Photoelectric equation. eV0=hν−ϕ, so ϕ=hν−eV0 (same hν for both metals).
Difference.
ϕA−ϕB=(hν−eV0A)−(hν−eV0B)=e(V0B−V0A)=e(1.6−0.4)=1.2 eV.
The magnitude of the difference in work functions is 1.2 eV.
✓Final answerThe correct option is (B).
- KEAM 2026Set eng-2026-04214 marksMCQQ.If a radiation of energy 5.2eV falls on the photosensitive surfaces of Mo and Ni, they emit photoelectrons with maximum kinetic energy of 0.5eV and 1eV, respectively. Then the work function of (A) Mo is 2.6eV (B) Ni is 6.2eV (C) Mo is 6.2eV (D) Ni is 4.2eV (E) Mo is 4.2eV
›Reveal solutionSolution
Photoelectric equation ϕ=E−KEmax: Ni =4.2eV.
Using KEmax=E−ϕ:
ϕMo=5.2−0.5=4.7eV,ϕNi=5.2−1=4.2eV.
Among the options, only “Ni is 4.2eV” is correct.
✓Final answerThe correct option is (D).
- KEAM 2026Set eng-2026-04224 marksMCQQ.If the energy of the incident radiation on a metal is increased by 10 %, the kinetic energy of the emitted photoelectrons increases from 0.5eV to 0.75 eV, then the work function of the metal is (A) 2 eV (B) 1.5 eV (C) 1 eV (D) 2.5 eV (E) 1.8 eV
›Reveal solutionSolution
Einstein's equation E=W+KE. Increasing E by 10% raises KE from 0.5 to 0.75 eV; solving 1.1(W+0.5)=W+0.75 gives W=2 eV.
Initially E=W+0.5. When the incident energy is increased by 10%:
1.1E=W+0.75.
Substituting E=W+0.5:
1.1(W+0.5)=W+0.75
1.1W+0.55=W+0.75
0.1W=0.20⇒W=2 eV.
✓Final answerThe correct option is (A).
- KEAM 2026Set pha-2026-0419F4 marksMCQQ.The work function of a material that has the threshold frequency of 5×1014 Hz is (h=6.626×10−34 Js) (A) 3.09 eV (B) 5.35 eV (C) 4.14 eV (D) 2.07 eV (E) 1.03 eV
›Reveal solutionSolution
Work function ϕ0=hν0=3.313×10−19 J≈2.07 eV.
The work function equals Planck's constant times the threshold frequency:
ϕ0=hν0=(6.626×10−34J s)(5×1014Hz)=3.313×10−19J.
Converting to electron-volts (1eV=1.6×10−19 J):
ϕ0=1.6×10−193.313×10−19≈2.07eV.
✓Final answerThe correct option is (D).
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.If the stopping potential in a photoelectric experiment is measured to be 1.82 V, the maximum speed of the emitted electrons, in ms−1, is (mass of the electron = 9.1×10−31 kg) (A) 8.0×105 (B) 2.3×105 (C) 3.0×105 (D) 7.3×106 (E) 5.3×1011
›Reveal solutionSolution
eV0=21mv2; v=2eV0/m=2(1.6×10−19)(1.82)/9.1×10−31≈8.0×105 m/s.
The stopping potential equals the maximum kinetic energy of the photoelectrons:
eV0=21mvmax2⇒vmax=m2eV0.
Substituting e=1.6×10−19C, V0=1.82V, m=9.1×10−31kg:
vmax=9.1×10−312(1.6×10−19)(1.82)=6.4×1011≈8.0×105m/s.
✓Final answerThe correct option is (A).
- KEAM 2025Set eng-2025-04234 marksMCQQ.Pick out the INCORRECT statement from the following : In photoelectric phenomenon, (A) the value of stopping potential is the same for radiations of all frequencies (B) the stopping potential is more negative for the incident radiation of higher frequency (C) the value of saturation current depends on the intensity of incident radiation (D) the value of saturation current is independent of frequency of incident radiation (E) the emission of electrons is instantaneous
›Reveal solutionSolution
The incorrect statement is that the stopping potential is the same for all frequencies.
Concept and Intuition
Einstein's photoelectric equation gives eV_0 = h*nu - phi, so the stopping potential increases with frequency. Saturation current depends on intensity, not frequency, and emission is instantaneous. Statement (A) contradicts the frequency dependence of stopping potential and is therefore the false one.
Step-by-Step Solution
- eV_0 = h*nu - phi, so V_0 depends on frequency.
- Hence (A) 'same for all frequencies' is wrong.
- (B), (C), (D), (E) are all correct features of the photoelectric effect.
Common Mistakes
- Confusing saturation current (intensity-dependent) with stopping potential (frequency-dependent).
✓Final answerThe correct option is (A) — the value of stopping potential is the same for radiations of all frequencies.
ANSWER: A
- KEAM 2025Set eng-2025-04254 marksMCQQ.If light waves of wavelengths λ and λ/3 are incident on the surface of a material, photoelectrons are emitted with maximum kinetic energy E and 4E respectively, then the work function of the material is (A) 2λhc (B) 3λhc (C) λhc (D) 2λ3hc (E) λ2hc
›Reveal solutionSolution
Apply Einstein's photoelectric equation to both wavelengths and eliminate E: the work function comes out as hc/3λ.
Einstein's equation KE=λhc−ϕ:
E=λhc−ϕ(1)
4E=λ/3hc−ϕ=λ3hc−ϕ(2)
Subtracting (1) from (2):
3E=λ3hc−λhc=λ2hc ⇒ E=3λ2hc.
Substituting back into (1):
ϕ=λhc−E=λhc−3λ2hc=3λhc.
✓Final answerThe correct option is (B).
- KEAM 2025Set eng-2025-04264 marksMCQQ.Which one of the following statements is INCORRECT? In photoelectric effect (A) Threshold frequency is different for different metals (B) The same metal gives same response to light of different wavelengths (C) The emission of photoelectrons is an instantaneous process (D) Above the threshold frequency the number of photoelectrons emitted per sec is directly proportional to the intensity of incident radiation (E) The maximum K.E. of the photoelectrons is independent of the intensity of incident radiation
›Reveal solutionSolution
A metal's photoelectric response depends strongly on the wavelength/frequency of light, so the claim that it responds identically to different wavelengths is false.
In the photoelectric effect, emission and the maximum kinetic energy of photoelectrons depend on the frequency (wavelength) of the incident light: KEmax=hν−ϕ0.
Statements (A), (C), (D) and (E) are all correct standard features of the photoelectric effect.
Statement (B), that the same metal gives the same response to light of different wavelengths, is INCORRECT, because different wavelengths have different photon energies and hence produce different responses (and below threshold, no emission at all).
✓Final answerThe correct option is (B).
- KEAM 2025Set eng-2025-04274 marksMCQQ.The plot of maximum kinetic energy of photo-electrons to the energy of the incident photon above its threshold frequency on a photo-sensitive material of work function φ is (A) an oblique straight line with a positive slope. (B) an oblique straight line with a negative slope. (C) an oblique straight line passing through the origin. (D) an exponential curve. (E) a polynomial curve of order 2.
›Reveal solutionSolution
Einstein's equation Kmax=hν−φ makes maximum KE a straight line in photon energy hν with positive slope +1.
Einstein's photoelectric equation is:
Kmax=hν−φ,
where hν is the incident photon energy and φ the work function. Plotting Kmax (y-axis) against the photon energy E=hν (x-axis) gives a straight line of slope +1 (positive) and intercept −φ on the KE axis.
Above the threshold frequency Kmax>0, so the graph is an oblique (inclined) straight line rising with a positive slope. It does not pass through the origin because of the −φ intercept; it crosses the energy axis at E=φ.
✓Final answerThe correct option is (A).
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