Consider a beam of electrons (each electron with energy E₀) incident on a metal surface kept in an evacuated c

NCERT Exemplar · Q3

E_0

(a) no electrons will be emitted as only photons can emit electrons.

(b) electrons can be emitted but all with an energy,

E_0

E_0 - \phi

(

\phi

is the work function).

(d) electrons can be emitted with any energy, with a maximum of

E_0

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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}

Step 2: Subtract work function

K_{\text{max}} = 2.48 - 2.0 = 0.48 \ \text{eV}

Step 3: Convert to joules if needed

0.48 \ \text{eV} \times 1.6 \times 10^{-19} = 7.68 \times 10^{-20} \ \text{J}

The electron escapes with this much kinetic energy.

Common Mistake to Avoid

Watch out

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, $\phi$ .

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

Quantity	Formula	Why it holds
Photon energy	$E = hf$	Light is quantized (Planck-Einstein)
Work function	$\phi = hf_0$	Minimum energy to escape at threshold
Max kinetic energy	$K_{\text{max}} = hf - \phi$	Energy conservation per photon-electron
Stopping potential	$eV_s = hf - \phi$	Electric work balances kinetic energy
Threshold frequency	$f_0 = \phi/h$	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.

An incident electron beam ejects electrons by collision (secondary emission); the freed electrons come out with a spread of energies whose maximum is $E_0-\phi$ — 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 $E_0$ . 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 $E_0$ .
Energy that must be paid. To escape the metal, the freed electron must spend at least the work function $\phi$ .
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:

K_{\max} = E_0 - \phi.

So the emitted electrons are not mono-energetic; they form a continuous distribution up to $E_0-\phi$ .

Why the other options fail

(A) "No electrons emitted" is wrong: an energetic electron beam does eject electrons by collision.
(B) "All with energy $E_0$ " is wrong: the incident electron loses part of its energy in the collision, and the escaping electron also pays $\phi$ .
(D) "Maximum of $E_0$ " ignores the work function that must be spent to leave the surface.

✓Final answer

E_0-\phi

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 $E_{\text{photon}} = hf = \phi + K_{\max}$ 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 $\phi$ — to leave the metal surface. So the maximum possible kinetic energy of an emitted electron is always

K_{\max} = (\text{energy available to the electron}) - \phi

Here the energy available is the full incident energy $E_0$ of the electron doing the knocking, so $K_{\max} = E_0 - \phi$ .

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 $\phi$ ignores that escape always has a cost.

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 20 on this concept.

GUJCET 2026Set x1 markMCQ
$V_0$
›Reveal solutionSolution

Einstein's equation gives $V_0=\frac{h}{e}\nu-\frac{\phi}{e}$ ; the slope is $\frac{h}{e}$ .

Einstein's photoelectric equation with stopping potential:
$eV_0=h\nu-\phi$
Dividing by $e$ :
$V_0=\frac{h}{e}\,\nu-\frac{\phi}{e}$
Comparing with a straight line $y=mx+c$ , the slope is
$m=\frac{h}{e}$
✓Final answer
$\frac{h}{e}$

ANSWER: (D)
GUJCET 2026Set x1 markMCQ
$Q. The photoelectric cut-off voltage in a certain experiment is 1.5 V. The kinetic energy of photoelectrons emitted will be ______. (A) 1.5 J (B) 1.5 eV (C) 2.4 eV (D) 2.4 J$
›Reveal solutionSolution

Max KE $=eV_0=1.5\text{ eV}$ (numerically equal to the stopping voltage in eV).

The cut-off (stopping) potential $V_0$ relates to the maximum kinetic energy of photoelectrons by
$K_{max}=eV_0$
With $V_0=1.5\text{ V}$ :
$K_{max}=e\times1.5\text{ V}=1.5\text{ eV}$
The energy in electron-volts is numerically equal to the stopping voltage in volts.

✓Final answer
$Option (B) 1.5 eV$

ANSWER: (B)
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. Slope of the graph of stopping potential (V_0) vs frequency (nu) is ___.$
(a) zero
(b) phi_0 / e
(c) h / e
(d) phi_0
›Reveal solutionSolution

Einstein's photoelectric equation, written in terms of the stopping potential, is a straight line in V_0 vs nu with slope h/e - this is how h/e was experimentally measured.

Einstein's equation: e V_0 = h nu - phi_0

Divide by e: V_0 = (h/e) nu - phi_0/e

This is of the form y = mx + c with y = V_0, x = nu, slope m = h/e, and intercept -phi_0/e (work function term).

✓Final answer
$(c) h/e.$
GUJCET 2025Set 031 markMCQ
$10^{9}$
›Reveal solutionSolution

Pulling electrons out of a metal by an external field (field/cold emission) needs a very strong field of order $10^8\ \text{V/m}$ , i.e. $10^6\ \text{V/cm}$ .

Concept. Electrons are bound to a metal by the work function. To rip them out purely by an electric field (field emission), the surface field must be extremely large, of order $10^8\ \text{V m}^{-1}$ (NCERT).

Unit conversion. $1\ \text{V/m} = 10^{-2}\ \text{V/cm}$ , so
$10^8\ \text{V/m} = 10^8 \times 10^{-2}\ \text{V/cm} = 10^6\ \text{V/cm}.$
✓Final answer
$10^6$

ANSWER: (B)
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Which phenomena cannot be explained by wave theory of light?$
(a) Interference
(b) Polarisation
(c) Diffraction
(d) Photo-electric effect
›Reveal solutionSolution

Interference, diffraction and polarisation are wave phenomena, fully explained by the wave theory of light; the photoelectric effect cannot be.

Wave theory predicts that photoelectric emission should depend on intensity (not frequency) and should show a time lag at low intensity — neither is observed. Only Einstein's photon picture (E = hν per photon, instantaneous energy transfer, existence of a threshold frequency) explains the photoelectric effect correctly.

✓Final answer
$(d) Photo-electric effect.$
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. The work function of Caesium is 2.14 eV. Find the threshold cut-off frequency for Caesium. [h = 6.63 x 10^-34 Js]$
(a) 3.22 x 10^33 Hz
(b) 3.22 x 10^14 Hz
(c) 5.16 x 10^15 Hz
(d) 5.16 x 10^14 Hz
›Reveal solutionSolution

The threshold frequency is the minimum frequency needed to just overcome the work function: ν0 = W0/h.

W0 = 2.14 eV = 2.14 × 1.6 × 10⁻¹⁹ = 3.424 × 10⁻¹⁹ J.

ν0 = W0/h = 3.424 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.164 × 10¹⁴ Hz.

✓Final answer
$(d) 5.16 \times 10¹⁴ Hz.$
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Which condition is satisfied for photoelectric effect in the metal given below?$
(a) Energy of incident photon (hν) is lesser than work function (φ0) of metal
(b) Wavelength of incident light (λ) is greater than threshold wavelength (λ0) of metal
(c) Frequency of incident light (ν) is greater than threshold frequency (ν0) of metal
(d) λ > hc/φ0
›Reveal solutionSolution

Photoelectric emission occurs only when the incident photon's energy exceeds the metal's work function, i.e. when ν > ν0.

Since E = hν and W0 = hν0, the emission condition hν ≥ W0 is equivalent to ν ≥ ν0. Equivalently, in terms of wavelength, λ must be less than the threshold wavelength λ0 (not greater, ruling out option b) — the other options either invert the inequality or use λ incorrectly.

✓Final answer
$(c) Frequency of incident light greater than threshold frequency.$
GUJCET 2024Set 131 markMCQ
$10^4$
›Reveal solutionSolution

To pull electrons out of a metal purely by a field (field emission), the surface field must be about $10^8\ \text{V/m}$ .

Concept. Overcoming the work function by an external field alone requires a very strong field, of order $10^8\ \text{V m}^{-1}$ (NCERT, field/cold emission).

✓Final answer
$10^8$

ANSWER: (D)
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. In the case of Photoelectric effect, on increasing the frequency of incident light, ___.$
(a) Photoelectric current increases
(b) Photoelectric current decreases
(c) Stopping potential increases
(d) Stopping potential decreases
›Reveal solutionSolution

Photoelectric current depends on the intensity of light (number of photons/sec), while the maximum kinetic energy (and hence stopping potential) depends on the frequency of light.

As frequency increases (intensity held fixed), each photon carries more energy (E = hf), so the maximum kinetic energy of ejected photoelectrons, KEmax = hf − φ0, increases. Since eV0 = KEmax, the stopping potential V0 increases linearly with frequency. Photoelectric current itself does not depend on frequency (only on the number of incident photons, i.e. intensity), so options (a) and (b) are incorrect.

✓Final answer
$(c) Stopping potential increases.$
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. Threshold frequency of which of the following metal does not lie in the ultraviolet region. (In case of photoelectric effect)$
(a) Zinc
(b) Cadmium
(c) Magnesium
(d) Sodium
›Reveal solutionSolution

Alkali metals (Na, K, Rb, Cs) have low work functions and show the photoelectric effect even with visible light, whereas metals like zinc, cadmium and magnesium require ultraviolet light.

Among the given options, zinc, cadmium and magnesium all have relatively higher work functions requiring UV radiation to eject photoelectrons. Sodium, with a much lower work function, has a threshold frequency that falls within the visible spectrum rather than the ultraviolet.

✓Final answer
$(d) Sodium.$
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Variation of stopping potential V_0 with frequency (nu) of incident radiation for a given photosensitive material is straight line. [frequency (nu) of incident radiation is greater than threshold frequency (nu)]. The slope of this line is ___.$
(a) h/e
(b) h/nu
(c) phi_0/h
(d) e/V_0
›Reveal solutionSolution

Rearranging the photoelectric equation gives V_0 = (h/e) nu - phi_0/e; the slope of V_0 versus nu is h/e.

Stopping potential relation: e V_0 = h*nu - phi_0.

Divide by e: V_0 = (h/e) nu - (phi_0/e).

This is a straight line of slope h/e (and intercept -phi_0/e). Millikan used exactly this to measure Planck's constant.

✓Final answer
$(a) h/e.$
GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. For the photoelectric effect of a metal, the slope of the graph of stopping potential (V0) versus the frequency (ν) of the incident light is ______.$
(a) e / h
(b) h / e
(c) h
(d) h / 2π
›Reveal solutionSolution

Rearranging Einstein's photoelectric equation into the form V0 = (h/e)ν - φ0/e shows the graph's slope is h/e.

Einstein's photoelectric equation: $eV_0 = h\nu - \phi_0$

Dividing by e: $V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}$

This is a straight line in $V_0$ vs $\nu$ , with slope $\dfrac{h}{e}$ and y-intercept $-\dfrac{\phi_0}{e}$ .

✓Final answer
$(b) h/e.$

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Q.Consider a beam of electrons (each electron with energy E0E_0E0​) incident on a metal surface kept in an evacuated chamber. Then

Setting up the physics

How much energy can an ejected electron have?

Why the other options fail

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Q.Consider a beam of electrons (each electron with energy $E_0$ ) incident on a metal surface kept in an evacuated chamber. Then

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