Q.How much positive and negative charge is there in a cup of water? (Assume the mass of one cup of water to be 250g.)
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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 …
Water is electrically neutral, but each molecule packs equal amounts of positive (protons) and negative (electrons) charge. We count one kind; the other is equal in magnitude.
- Moles of water. Molar mass of H2O=18g/mol, so 250g is
n=18250=13.9mol.
- Molecules. N=nNA=13.9×6.022×1023=8.36×1024.
- Protons (or electrons) per molecule. H2O has 2(1)+8=10 of each.
- Total positive charge. …
Counting the 10 protons (and 10 electrons) in every water molecule, a 250g cup contains about 1.34×107C of positive charge and an equal magnitude of negative charge — they cancel, which is why water is neutral.
This question is about the charge that matter contains, not any net charge it carries. Every neutral molecule holds equal positive and negative charge; the task is to work out how large each of those is for a cup of water.
Step 1 — Number of moles.
The molar mass of water is M=2(1)+16=18g/mol, so
n=Mm=18250=13.9mol.
Step 2 — Number of molecules.
N=nNA=13.9×(6.022×1023)=8.36×1024 molecules.
Step 3 — Charges per molecule.
Each H2O has 2 hydrogen atoms (1 proton each) and 1 oxygen atom (8 protons), i.e. 10 protons and, being neutral, 10 electrons.
Step 4 — Total positive charge.
Total protons =10N=8.36×1025, so …
Method: Atomic Charge Counting via Avogadro's Number
This method uses the fact that every atom in water is electrically neutral overall, but contains equal amounts of positive charge (in protons) and negative charge (in electrons). We count the total number of each type of charge carrier.
Step 1: Find the number of water molecules
- Molar mass of water (H2O): 2×1+16=18 g/mol
- Number of moles in 250 g of water: n=18250≈13.89 mol
- Using Avogadro's number (NA=6.022×1023 mol−1): Nmolecules=n×NA≈13.89×6.022×1023≈8.36×1024 molecules
Step 2: Count protons and electrons per molecule
Each water molecule has:
- 10 electrons (8 from oxygen + 1 from each hydrogen)
- 10 protons (8 from oxygen + 1 from each hydrogen)
So total number of protons = total number of electrons:
Nprotons=Nelectrons=10×8.36×1024=8.36×1025
Step 3: Compute the total charge
Charge of one proton: +e=+1.602×10−19 C …
Common Mistakes: Charge in a Cup of Water
Students often make predictable errors on this question. Here are the most common ones — and how to avoid each.
1. Forgetting that water is electrically neutral
Mistake:
Students calculate only the number of electrons or protons, then report that as the net charge.
Why it’s wrong:
A cup of water has equal numbers of protons and electrons. The net charge is zero. The question asks for the amount of positive and negative charge separately, not the net charge.
How to avoid:
Always start by stating: “Water is neutral — total positive charge equals total negative charge.” Then calculate the magnitude of one type.
2. Using the wrong molecular formula or atomic numbers
Mistake:
Treating water as H2O but forgetting that:
- Hydrogen has 1 proton (atomic number Z=1)
- Oxygen has 8 protons (Z=8)
So total protons per water molecule = 1+1+8=10.
How to avoid:
Write it down explicitly:
Protons per H2O=2×1+1×8=10
3. Confusing mass with number of molecules
Mistake:
Using 250g directly without converting to moles.
How to avoid:
Always use the molar mass:
- Molar mass of water = 18g/mol
- Number of moles = 18250≈13.89mol
- Number of molecules = 13.89×6.022×1023
4. Using the wrong value for elementary charge
Mistake:
Using e=1.6×10−19C but forgetting that proton charge is +e and electron charge is −e.
How to avoid:
Write clearly:
- Charge of one proton = +1.6×10−19C
- Charge of one electron = −1.6×10−19C
5. Reporting the answer with wrong sign or units …
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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