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.

JKBOSE Class 12 Annual Regular Examination 2023Set ANNUAL1 mark
$Q. What is the effect of decrease in wavelength of incident light on the velocity of photoelectrons ?$
›Reveal solutionSolution

A shorter wavelength means a higher-energy photon, so by Einstein's photoelectric equation the maximum kinetic energy - and hence the maximum velocity - of the emitted photoelectrons increases.

By Einstein's photoelectric equation, when light of frequency nu (wavelength lambda, with nu = c/lambda) falls on a metal surface of work function phi0, the maximum kinetic energy of an emitted photoelectron is

KE_max = hnu - phi0 = (hc/lambda) - phi0 = (1/2) m v_max^2.

Here h is Planck's constant, c the speed of light, m the electron mass. If the wavelength lambda of the incident light is decreased (keeping the metal, and hence phi0, fixed), the photon energy hc/lambda increases. Since phi0 is a fixed property of the metal, KE_max = (hc/lambda) - phi0 increases as lambda decreases.

Because KE_max = (1/2) m v_max^2, an increase in KE_max means the maximum velocity v_max of the emitted photoelectrons also increases:

v_max = sqrt[2(hc/lambda - phi0)/m].

So a decrease in wavelength increases the maximum velocity of the photoelectrons, provided lambda remains below the threshold wavelength lambda0 = hc/phi0 so that photoemission occurs at all. (Note that decreasing lambda does NOT change the number of photoelectrons emitted per second - that depends on intensity, not wavelength.)

✓Final answer
$Decreasing the wavelength of incident light increases the maximum kinetic energy and hence the maximum velocity of the emitted photoelectrons, since KE_max = hc/lambda - phi0 increases as lambda decreases.$
JKBOSE Class 12 Annual Regular Examination 2022Set SZ1 mark
$Q. Define threshold frequency.$
›Reveal solutionSolution

Threshold frequency is the minimum light frequency needed to just eject an electron from a metal surface, related to the work function by $W_0 = h\nu_0$ .

When light of a certain frequency falls on a metal surface, photoelectrons are emitted only if the frequency of the incident light is high enough to supply at least the minimum energy needed to free an electron from the metal (the work function $W_0$ ).

The threshold frequency, denoted $\nu_0$ , is defined as the minimum frequency of incident radiation which can just cause photoelectric emission from a given metal surface, with zero kinetic energy given to the emitted electron. For frequencies below $\nu_0$ , no photoelectrons are emitted at all, regardless of how intense (bright) the incident light is — only the number of electrons emitted increases with intensity, not whether emission occurs.

It is related to the work function of the metal by:

$W_0 = h u_0$

where $h$ is Planck's constant. Threshold frequency is characteristic of the metal (different metals have different threshold frequencies/work functions).

✓Final answer
$\nu_0$
JKBOSE Class 12 Annual Regular Examination 2021Set SZ1 mark
$Q. Define Work Function.$
›Reveal solutionSolution

Work function is the minimum binding energy that must be supplied to free the most loosely bound electron from a metal surface.

In a metal, free electrons are held inside by the attractive forces of positive ions; energy is needed to pull an electron out through the surface.

The work function $\phi_0$ of a metal is defined as the minimum energy required to just liberate an electron from the metal surface, so that it can escape with zero kinetic energy.

It is related to the threshold frequency $\nu_0$ (the minimum frequency of incident light that can cause photoemission) by

$\phi_0 = h u_0 = \frac{hc}{\lambda_0}$

where $h$ is Planck's constant and $\lambda_0$ is the threshold wavelength.

Work function is usually expressed in electron-volts (eV) and varies from metal to metal — it is lowest for alkali metals like caesium (~2.14 eV) and higher for metals like platinum.

✓Final answer
$\phi_0 = h\nu_0$
JKBOSE Class 12 Annual Regular Examination 2018Set ANNUAL1 markMCQ
$Q. Work function is measured in :$
(a) Electrovolt
(b) Volt
(c) Joul/sec
(d) Watt
›Reveal solutionSolution

Work function is an energy, and by convention is expressed in electronvolts because joules would be inconveniently small numbers.

Work function ( $\phi_0$ ) is the minimum energy required to just liberate an electron from the surface of a metal (with zero kinetic energy). Being an energy, its SI unit is the joule, but because typical work function values are extremely small in joules (of order $10^{-19}$ J), it is conventionally expressed in electronvolts (eV), where $1\ \text{eV} = 1.6\times10^{-19}$ J — a far more convenient scale (typical metals have $\phi_0 \approx 2$ –5 eV). Volt and watt are units of potential and power respectively, not energy, and "Joule/sec" is watt (power), not energy — so neither is appropriate for work function.

✓Final answer
$(a) Electronvolt.$

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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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