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 21 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.The work function of a material is 6.6 eV. Then, the threshold wavelength of the metal is approximately (Take h=6.6×10−34 J.s) (A) 108 nm (B) 188 nm (C) 208 nm (D) 228 nm (E) 250 nm
›Reveal solutionSolution
Threshold wavelength satisfies W=hc/λ0. With W=6.6 eV this gives about 188 nm.
Convert the work function to joules:
W=6.6 eV=6.6×1.6×10−19=1.056×10−18 J.
Threshold wavelength: …
- KEAM 2026Set eng-2026-04184 marksMCQQ.Two light rays of wavelength λ and 4λ incident on the surface of a photo sensitive material emit electrons with max kinetic energy E and 6E respectively. The work function of the material is (h = Planck's constant, c = velocity of light in free space) (A) λhc (B) 2λhc (C) 5λ2hc (D) 5λ3hc (E) 3λhc
›Reveal solutionSolution
Writing Einstein's equation for both wavelengths and eliminating E gives the work function W=5λ2hc.
Einstein's photoelectric equation, KE=λhc−W:
For wavelength λ:
E=λhc−W(1)
For wavelength λ/4 (photon energy 4hc/λ):
6E=λ4hc−W(2)
Subtract (1) from (2): …
- KEAM 2026Set eng-2026-04194 marksMCQQ.If the threshold wavelengths of two metals are in the ratio 1:3, then the work functions of these metals are in the ratio (A) 1:3 (B) 2:1 (C) 3:1 (D) 1:2 (E) 3:2
›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
ϕ=λ0hc⇒ϕ∝λ01.
Given λ0,1:λ0,2=1:3, …
- KEAM 2026Set eng-2026-04204 marksMCQQ.Stopping potentials for the metals A and B are 0.4 V and 1.6 V, respectively. When illuminated by same light, the difference in their work functions is: (A) 2.0 eV (B) 1.2 eV (C) 6.4 eV (D) 0.4 eV (E) 2.0 V
›Reveal solutionSolution
With identical incident light, the work-function difference equals e times the stopping-potential difference: 1.2 eV.
Photoelectric equation. eV0=hν−ϕ, so ϕ=hν−eV0 (same hν for both metals).
Difference. …
- KEAM 2026Set eng-2026-04214 marksMCQQ.If a radiation of energy 5.2eV falls on the photosensitive surfaces of Mo and Ni, they emit photoelectrons with maximum kinetic energy of 0.5eV and 1eV, respectively. Then the work function of (A) Mo is 2.6eV (B) Ni is 6.2eV (C) Mo is 6.2eV (D) Ni is 4.2eV (E) Mo is 4.2eV
›Reveal solutionSolution
Photoelectric equation ϕ=E−KEmax: Ni =4.2eV.
Using KEmax=E−ϕ:
ϕMo=5.2−0.5=4.7eV,ϕNi=5.2−1=4.2eV. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.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: …
- KEAM 2026Set pha-2026-0419F4 marksMCQQ.The work function of a material that has the threshold frequency of 5×1014 Hz is (h=6.626×10−34 Js) (A) 3.09 eV (B) 5.35 eV (C) 4.14 eV (D) 2.07 eV (E) 1.03 eV
›Reveal solutionSolution
Work function ϕ0=hν0=3.313×10−19 J≈2.07 eV.
The work function equals Planck's constant times the threshold frequency:
ϕ0=hν0=(6.626×10−34J s)(5×1014Hz)=3.313×10−19J. …
- KEAM 2026Set pha-2026-0420F4 marksMCQQ.If the stopping potential in a photoelectric experiment is measured to be 1.82 V, the maximum speed of the emitted electrons, in ms−1, is (mass of the electron = 9.1×10−31 kg) (A) 8.0×105 (B) 2.3×105 (C) 3.0×105 (D) 7.3×106 (E) 5.3×1011
›Reveal solutionSolution
eV0=21mv2; v=2eV0/m=2(1.6×10−19)(1.82)/9.1×10−31≈8.0×105 m/s.
The stopping potential equals the maximum kinetic energy of the photoelectrons:
eV0=21mvmax2⇒vmax=m2eV0.
Substituting e=1.6×10−19C, V0=1.82V, m=9.1×10−31kg: …
- KEAM 2025Set eng-2025-04234 marksMCQQ.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. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.If light waves of wavelengths λ and λ/3 are incident on the surface of a material, photoelectrons are emitted with maximum kinetic energy E and 4E respectively, then the work function of the material is (A) 2λhc (B) 3λhc (C) λhc (D) 2λ3hc (E) λ2hc
›Reveal solutionSolution
Apply Einstein's photoelectric equation to both wavelengths and eliminate E: the work function comes out as hc/3λ.
Einstein's equation KE=λhc−ϕ:
E=λhc−ϕ(1)
4E=λ/3hc−ϕ=λ3hc−ϕ(2)
Subtracting (1) from (2): …
- KEAM 2025Set eng-2025-04264 marksMCQQ.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: KEmax=hν−ϕ0.
Statements (A), (C), (D) and (E) are all correct standard features of the photoelectric effect. …
- KEAM 2025Set eng-2025-04274 marksMCQQ.The plot of maximum kinetic energy of photo-electrons to the energy of the incident photon above its threshold frequency on a photo-sensitive material of work function φ is (A) an oblique straight line with a positive slope. (B) an oblique straight line with a negative slope. (C) an oblique straight line passing through the origin. (D) an exponential curve. (E) a polynomial curve of order 2.
›Reveal solutionSolution
Einstein's equation Kmax=hν−φ makes maximum KE a straight line in photon energy hν with positive slope +1.
Einstein's photoelectric equation is:
Kmax=hν−φ,
where hν is the incident photon energy and φ the work function. Plotting Kmax (y-axis) against the photon energy E=hν (x-axis) gives a straight line of slope +1 (positive) and intercept −φ on the KE axis. …
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