Q.(a) Explain the meaning of the statement 'electric charge of a body is quantised'.
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🔒 Start your 14-day free trial to unlock the full solution →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 …
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 …
- The statement means that electric charge on any body is always an integer multiple of the elementary charge e=1.6×10−19C. Charge cannot exist in arbitrary fractions of e — it comes in discrete packets. If a body has n excess or deficit electrons, its net charge is q=±ne, where n is an integer.
- At the macroscopic scale, the total charge involved is typically of the order of microcoulombs (μC) or more. For example, 1μC corresponds to about 6.25×1012 electrons. …
Electric charge is quantised, meaning it exists only in integer multiples of the elementary charge e=1.6×10−19C. At macroscopic scales, the number of charge carriers is so enormous that the discrete jumps become negligible, and charge behaves as if it were continuous.
(a) What "quantised" means for electric charge
The statement "electric charge of a body is quantised" means that any observable charge Q on an object is always an integer multiple of a fundamental, smallest unit of charge. That unit is the magnitude of the charge on a single electron or proton, denoted by e:
e=1.602×10−19C
So if a body has a net charge Q, it must satisfy:
Q=±ne,where n=0,1,2,3,…
You cannot have, say, 0.5e or 1.7e on a body in isolation. Charge comes in discrete packets — it is not a continuous fluid that can be divided arbitrarily. This was first demonstrated convincingly by Millikan's oil drop experiment, which showed that every measured charge was a multiple of e.
The quantisation of charge is a fundamental law of nature. It arises because matter is made of electrons and protons, each carrying exactly ±e. Any transfer of charge involves moving whole electrons or protons — you cannot transfer a fraction of an electron.
(b) Why we ignore quantisation at macroscopic scales
When dealing with macroscopic (large-scale) charges — say, a charged metal sphere carrying 1μC — the number of excess electrons (or protons) is enormous. Let's calculate:
n=eQ=1.6×10−19C1×10−6C≈6.25×1012
That's over six trillion elementary charges. Now, if you add or remove even a few billion electrons, the change in charge is:
ΔQ=(109)×(1.6×10−19)=1.6×10−10C
This is 0.16 nanocoloumbs — far below the sensitivity of most macroscopic measuring instruments. The relative jump between allowed charge values is:
Qe≈10−61.6×10−19=1.6×10−13 …
Method: Quantisation of Charge — Explanation & Justification
This is a concept-based reasoning method, not a calculation. The steps below apply to both parts (a) and (b).
(a) Meaning of “electric charge of a body is quantised”
Step 1 – State the fundamental idea
Electric charge cannot exist in arbitrary amounts. It always comes in discrete packets.
Step 2 – Identify the smallest unit
The smallest possible free charge is the magnitude of the electron’s charge:
e=1.6×10−19C
Step 3 – Express the quantisation condition
Any observable charge Q on a body must be an integer multiple of e:
Q=±newhere n=0,1,2,3,…
Step 4 – Give the physical reason
This arises because charge is carried by electrons and protons — indivisible particles (under normal conditions). You cannot have half an electron.
Key result: Charge is quantised means Q=ne, with n an integer.
(b) Why quantisation is ignored for macroscopic charges
Step 1 – Compare magnitudes
Macroscopic charges (e.g., on a comb, a capacitor plate) are typically of the order of microcoulombs (μC) or more.
1μC=10−6C
Step 2 – Find the number of electrons involved
Number of electrons in 1μC: …
Here are the common mistakes students make on this question about quantisation of charge in the Photoelectric Effect chapter, along with how to avoid each.
Mistake 1: Confusing "Quantisation" with "Conservation"
- The Error: Students often write that "charge is quantised" means "charge cannot be created or destroyed." This is the law of conservation of charge, not quantisation.
- Why it happens: Both concepts appear in the same chapter (Photoelectric Effect / Dual Nature), and the words sound similar.
- How to avoid: Remember the keyword: "packets" or "multiples."
- Quantisation = charge exists only in discrete packets (multiples of e).
- Conservation = total charge in an isolated system remains constant.
Mistake 2: Forgetting the Exact Value of the Fundamental Charge
- The Error: Stating that charge is quantised in multiples of "the charge of an electron" but not giving the numerical value, or writing the wrong value (e.g., 1.6×10−19 C is correct; 1.6×10−19 J is wrong).
- Why it happens: Rushing through the definition without memorising the constant.
- How to avoid: Always write the exact statement:
"Electric charge exists only in discrete packets which are integral multiples of the elementary charge e=1.6×10−19 C."
- Formula to memorise: q=±ne, where n=0,1,2,3,…
Mistake 3: Writing "Charge is Quantised" Without the Integral Multiple Condition
- The Error: Saying "charge is quantised" but not specifying that it must be an integer multiple of e. Some students write "any multiple" (implying fractional multiples like 0.5e are allowed).
- Why it happens: Overlooking the word "integral" in the NCERT definition.
- How to avoid: Explicitly state: q=±ne, where n is an integer (n=0,1,2,…). No fractional n is allowed.
Mistake 4: Forgetting the "Why" for Macroscopic Charges (Part b)
- The Error: Simply saying "because charges are large" without explaining why we can ignore quantisation.
- Why it happens: Students memorise the answer but don't understand the logic.
- How to avoid: Use the analogy of grains of sand:
- A single grain of sand is like e (tiny).
- A bucket of sand is like a macroscopic charge (huge).
- When you measure the bucket's mass, you don't count individual grains — the discreteness is negligible compared to the total.
- Key line: "Since e is very small (1.6×10−19 C), a macroscopic charge contains an enormous number of electrons (n≈1019 or more). The addition or removal of a few electrons causes an imperceptible change, so the charge appears continuous."
Mistake 5: Mixing Up "Quantisation of Charge" with "Quantisation of Energy" …
Showing the 12 most recent of 20 on this concept.
- GUJCET 2026Set x1 markMCQQ.In photoelectric effect the graph of stopping potential (V0) versus frequency (ν) is a straight line. The slope of this graph is ______. (A) h (B) he (C) eV0 (D) eh
›Reveal solutionSolution
Einstein's equation gives V0=ehν−eϕ; the slope is eh.
Einstein's photoelectric equation with stopping potential:
eV0=hν−ϕ
Dividing by e:
V0=ehν−eϕ …
- GUJCET 2026Set x1 markMCQQ.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 =eV0=1.5 eV (numerically equal to the stopping voltage in eV).
The cut-off (stopping) potential V0 relates to the maximum kinetic energy of photoelectrons by
Kmax=eV0
With V0=1.5 V:
Kmax=e×1.5 V=1.5 eV …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.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
…
- GUJCET 2025Set 031 markMCQQ.The minimum value of electric field required to pulled out electrons from a metal is approximately ______ V/cm. (A) 109 (B) 106 (C) 1010 (D) 108
›Reveal solutionSolution
Pulling electrons out of a metal by an external field (field/cold emission) needs a very strong field of order 108 V/m, i.e. 106 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 108 V m−1 (NCERT). …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.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 …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.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. …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.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 …
- GUJCET 2024Set 131 markMCQQ.To emit an electron from the metal, minimum electric field required is ________. (A) 104 Vm−1 (B) 106 Vm−1 (C) 105 Vm−1 (D) 108 Vm−1
›Reveal solutionSolution
To pull electrons out of a metal purely by a field (field emission), the surface field must be about 108 V/m. …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.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 …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.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.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.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).
…
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.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: eV0=hν−ϕ0
Dividing by e: V0=ehν−eϕ0
…
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