Consider a thin target ($10^{-2}\ \text{m}$ square, $10^{-3}\ \text{m}$ thickness) of sodium, which produces a photocurrent of $100\ \mu\text{A}$ when a light of intensity $100\ \text{W/m}^2$ ($\lambda = 660\ \text{nm}$) falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom. [Take density of Na $= 0.97\ \text{kg/m}^3$.]

Consider a thin target (10⁻²m square, 10⁻³m thickness) of sodium, which produces a photocurrent of 100muA when

NCERT Exemplar · Q19

10^{-2}\ \text{m}

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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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The number of sodium atoms in the target ( $\approx 2.5\times10^{18}$ ) far exceeds the number of photons arriving each second ( $\approx 3.3\times10^{16}$ ), so effectively every incident photon strikes an atom. The probability of producing a photoelectron per photon-atom strike is the ratio of photoelectrons emitted per second to photons incident per second: $P \approx 0.019$ (about $2\%$ ).

Solution

Photoelectrons emitted per second. The photocurrent is $i = 100\ \mu\text{A} = 10^{-4}\ \text{A}$ , so

n_e = \frac{i}{e} = \frac{10^{-4}}{1.6\times10^{-19}} = 6.25\times10^{14}\ \text{s}^{-1}.

Photons incident per second. The power intercepted by the target of area $A = (10^{-2})^2 = 10^{-4}\ \text{m}^2$ is

P_{\text{inc}} = (100\ \text{W/m}^2)(10^{-4}\ \text{m}^2) = 10^{-2}\ \text{W}.

Each photon carries

E_\gamma = \frac{hc}{\lambda} = \frac{(6.62\times10^{-34})(3\times10^8)}{660\times10^{-9}} = 3.0\times10^{-19}\ \text{J},

n_\gamma = \frac{P_{\text{inc}}}{E_\gamma} = \frac{10^{-2}}{3.0\times10^{-19}} = 3.3\times10^{16}\ \text{s}^{-1}.

Number of sodium atoms in the target. The volume is $V = A\times t = 10^{-4}\times10^{-3} = 10^{-7}\ \text{m}^3$ ; with $\rho = 0.97\ \text{kg/m}^3$ and molar mass $M = 23\times10^{-3}\ \text{kg/mol}$ , …

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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 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 $e$ :

$V_0=\frac{h}{e}\,\nu-\frac{\phi}{e}$ …
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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 $V_0=1.5\text{ V}$ :

$K_{max}=e\times1.5\text{ V}=1.5\text{ eV}$ …
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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

…

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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). …
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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 photo …

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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. …

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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) — th …

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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 $10^8\ \text{V/m}$ . …

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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. Photoel …

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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.

…

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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).

…

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

…

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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 …

Solution

Method: Computing Event Probabilities from Rate Ratios

Steps

Solution

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