In a photoelectric experiment carried out with two different metals A and B, the stopping potential $V_{stop}$ (in volts) is measured as a function of the frequency $\nu$ (in Hz) of the incident light. For each metal the measured points lie on a straight line, and the two straight lines are parallel to each other (they have the same slope). The line for metal A lies to the left of the line for metal B. Extrapolated down to zero stopping potential, the line for A meets the frequency axis at a threshold frequency of about $5\times10^{14}$ Hz, while the line for B meets it at about $10\times10^{14}$ Hz; each line then rises with the same slope (for instance, line A passes close to the point $\nu=10\times10^{14}$ Hz, $V_{stop}=2$ V, and line B passes close to the point $\nu=15\times10^{14}$ Hz, $V_{stop}=2$ V). (i) Which material, A or B, has the higher work function? (ii) Given that the electric charge of an electron is $1.6\times10^{-19}$ C, find the value of Planck's constant $h$ obtained from this experiment for both A and B, and comment on whether the result is consistent with Einstein's photoelectric theory.

In a photoelectric experiment carried out with two different metals A and B, the stopping potential Vstop (in

NCERT Exemplar · Q21

V_{stop}

(i) Which material, A or B, has the higher work function?

(ii) Given that the electric charge of an electron is

1.6\times10^{-19}

C, find the value of Planck's constant

h

obtained from this experiment for both A and B, and comment on whether the result is consistent with Einstein's photoelectric theory.

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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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Both metals give parallel straight lines of $V_{stop}$ versus $\nu$ , so they share the same slope $h/e$ . Metal B's line meets the frequency axis at the higher threshold frequency, so B has the larger work function. The common slope gives $h\approx6.4\times10^{-34}$ J·s for both metals, matching Einstein's constant — confirming his theory.

Concept — Einstein's photoelectric equation

The stopping potential satisfies

eV_{stop}=h\nu-\phi_0,

which rearranges to

V_{stop}=\frac{h}{e}\,\nu-\frac{\phi_0}{e}.

This is a straight line in the $(\nu,\,V_{stop})$ plane with slope $\dfrac{h}{e}$ and a frequency-axis intercept (where $V_{stop}=0$ ) at the threshold frequency $\nu_0=\dfrac{\phi_0}{h}$ .

(i) Which has the higher work function

The work function is $\phi_0=h\nu_0$ , so a larger threshold frequency (an intercept further to the right on the frequency axis) means a larger work function. Line B meets the axis at $\nu_{0B}\approx10\times10^{14}$ Hz while line A meets it at $\nu_{0A}\approx5\times10^{14}$ Hz, so

\phi_{0B}=h\nu_{0B}>\phi_{0A}=h\nu_{0A}.

Metal B has the higher work function.

(ii) Value of Planck's constant

The slope of each line is $h/e$ , so $h=e\times(\text{slope})$ .

For material A, using $V_{stop}=0$ at $\nu=5\times10^{14}$ Hz and $V_{stop}=2$ V at $\nu=10\times10^{14}$ Hz:

\text{slope}_A=\frac{2-0}{(10-5)\times10^{14}}=4\times10^{-15}\ \text{V·s},

h_A=e\cdot\text{slope}_A=(1.6\times10^{-19})(4\times10^{-15})=6.4\times10^{-34}\ \text{J·s}.

For material B, using $V_{stop}=0$ at $\nu=10\times10^{14}$ Hz and $V_{stop}=2$ V at $\nu=15\times10^{14}$ Hz:

\text{slope}_B=\frac{2-0}{(15-10)\times10^{14}}=4\times10^{-15}\ \text{V·s},

h_B=e\cdot\text{slope}_B=(1.6\times10^{-19})(4\times10^{-15})=6.4\times10^{-34}\ \text{J·s}.

Comment …

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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 21 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$h = 6.6 \times 10^{-34}$
›Reveal solutionSolution

Threshold wavelength satisfies $W = hc/\lambda_0$ . With $W = 6.6$ eV this gives about 188 nm.

Convert the work function to joules:
$W = 6.6\ \text{eV} = 6.6 \times 1.6\times10^{-19} = 1.056\times10^{-18}\ \text{J}.$
Threshold wavelength: …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\lambda$
›Reveal solutionSolution

Writing Einstein's equation for both wavelengths and eliminating $E$ gives the work function $W=\dfrac{2hc}{5\lambda}$ .

Einstein's photoelectric equation, $KE=\dfrac{hc}{\lambda}-W$ :

For wavelength $\lambda$ :
$E=\frac{hc}{\lambda}-W\quad(1)$
For wavelength $\lambda/4$ (photon energy $4hc/\lambda$ ):
$6E=\frac{4hc}{\lambda}-W\quad(2)$
Subtract (1) from (2): …
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KEAM 2026Set eng-2026-04194 marksMCQ
$1 : 3$
›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
$\phi=\frac{hc}{\lambda_0}\Rightarrow\phi\propto\frac1{\lambda_0}.$
Given $\lambda_{0,1}:\lambda_{0,2}=1:3$ , …
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KEAM 2026Set eng-2026-04204 marksMCQ
$2.0\ eV$
›Reveal solutionSolution

With identical incident light, the work-function difference equals $e$ times the stopping-potential difference: $1.2$ eV.

Photoelectric equation. $eV_0=h\nu-\phi$ , so $\phi=h\nu-eV_0$ (same $h\nu$ for both metals).

Difference. …
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KEAM 2026Set eng-2026-04214 marksMCQ
$5.2\,eV$
›Reveal solutionSolution

Photoelectric equation $\phi=E-KE_{max}$ : Ni $=4.2\,$ eV.

Using $KE_{max}=E-\phi$ :

$\phi_{Mo}=5.2-0.5=4.7\,eV,\qquad \phi_{Ni}=5.2-1=4.2\,eV.$ …
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KEAM 2026Set eng-2026-04224 marksMCQ
$Q. If the energy of the incident radiation on a metal is increased by 10 %, the kinetic energy of the emitted photoelectrons increases from 0.5eV to 0.75 eV, then the work function of the metal is (A) 2 eV (B) 1.5 eV (C) 1 eV (D) 2.5 eV (E) 1.8 eV$
›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 $E=W+0.5$ : …
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KEAM 2026Set pha-2026-0419F4 marksMCQ
$5\times10^{14}$
›Reveal solutionSolution

Work function $\phi_0 = h\nu_0 = 3.313\times10^{-19}\ \mathrm{J} \approx 2.07\ \mathrm{eV}$ .

The work function equals Planck's constant times the threshold frequency:

$\phi_0 = h\nu_0 = (6.626\times10^{-34}\\ \text{J s})(5\times10^{14}\\ \text{Hz}) = 3.313\times10^{-19}\\ \text{J}.$ …
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KEAM 2026Set pha-2026-0420F4 marksMCQ
$V$
›Reveal solutionSolution

$eV_0=\tfrac12 mv^2$ ; $v=\sqrt{2eV_0/m}=\sqrt{2(1.6\times10^{-19})(1.82)/9.1\times10^{-31}}\approx8.0\times10^5$ m/s.

The stopping potential equals the maximum kinetic energy of the photoelectrons:
$eV_0 = \tfrac{1}{2}mv_{max}^2 \Rightarrow v_{max} = \sqrt{\frac{2eV_0}{m}}.$
Substituting $e = 1.6\times10^{-19}\,\mathrm{C}$ , $V_0 = 1.82\,\mathrm{V}$ , $m = 9.1\times10^{-31}\,\mathrm{kg}$ : …
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KEAM 2025Set eng-2025-04234 marksMCQ
$Q. Pick out the INCORRECT statement from the following : In photoelectric phenomenon, (A) the value of stopping potential is the same for radiations of all frequencies (B) the stopping potential is more negative for the incident radiation of higher frequency (C) the value of saturation current depends on the intensity of incident radiation (D) the value of saturation current is independent of frequency of incident radiation (E) the emission of electrons is instantaneous$
›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. …
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KEAM 2025Set eng-2025-04254 marksMCQ
$\lambda$
›Reveal solutionSolution

Apply Einstein's photoelectric equation to both wavelengths and eliminate $E$ : the work function comes out as $hc/3\lambda$ .

Einstein's equation $KE=\dfrac{hc}{\lambda}-\phi$ :
$E=\dfrac{hc}{\lambda}-\phi \quad(1)$ $4E=\dfrac{hc}{\lambda/3}-\phi=\dfrac{3hc}{\lambda}-\phi \quad(2)$
Subtracting (1) from (2): …
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KEAM 2025Set eng-2025-04264 marksMCQ
$Q. Which one of the following statements is INCORRECT? In photoelectric effect (A) Threshold frequency is different for different metals (B) The same metal gives same response to light of different wavelengths (C) The emission of photoelectrons is an instantaneous process (D) Above the threshold frequency the number of photoelectrons emitted per sec is directly proportional to the intensity of incident radiation (E) The maximum K.E. of the photoelectrons is independent of the intensity of incident radiation$
›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: $KE_{max} = h\nu - \phi_0$ .

Statements (A), (C), (D) and (E) are all correct standard features of the photoelectric effect. …
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KEAM 2025Set eng-2025-04274 marksMCQ
$\varphi$
›Reveal solutionSolution

Einstein's equation $K_{max} = h\nu - \varphi$ makes maximum KE a straight line in photon energy $h\nu$ with positive slope $+1$ .

Einstein's photoelectric equation is:
$K_{max} = h\nu - \varphi,$
where $h\nu$ is the incident photon energy and $\varphi$ the work function. Plotting $K_{max}$ (y-axis) against the photon energy $E = h\nu$ (x-axis) gives a straight line of slope $+1$ (positive) and intercept $-\varphi$ on the KE axis. …
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Concept understanding — Photoelectric Effect

The Photoelectric Effect: When Light Knocks Electrons Loose

The Intuition

The Precise Statement

What Classical Physics Got Wrong

The Key Insight

A Worked Example

Why this formula?

Photoelectric Effect: Why the Key Formulas Hold

1. The Core Idea: Energy Conservation

Why only one electron?

2. The Photoelectric Equation

Why "maximum" kinetic energy?

3. The Stopping Potential Connection

Combining:

4. Why the Threshold Frequency Exists

Why intensity doesn't matter for ejection?

5. Why Kinetic Energy Depends on Frequency, Not Intensity

6. Summary of Key Relationships …

Concept — Einstein's photoelectric equation

(i) Which has the higher work function

(ii) Value of Planck's constant

Comment …

Method: Extracting Physical Constants from a Stopping-Potential vs Frequency Graph

Steps

Concept and Intuition

Step-by-Step Solution

Concept — Einstein's photoelectric equation

(i) Which has the higher work function

(ii) Value of Planck's constant

Comment …

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