Q.Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of 3.3 V. If photo electric emission begins in this metal at a frequency of 8×1014 Hz, calculate the work function (in eV) for this metal. OR Monochromatic light of frequency 6.0×1014 Hz is produced by a laser. The power emitted is 2.0×10−3 W. Calculate the
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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 stopping potential fixes the maximum kinetic energy (eV0=3.3eV) and the threshold frequency fixes the work function (W0=hu0); Einstein's equation then gives the incident frequency. …
W0=hu0=3.3eV; ν=(W0+eV0)/h≈1.6×1015Hz. (OR: E=hν=3.98×10−19J, n=P/E≈5.0×1015s−1.)
Concept. Einstein's photoelectric equation: hν=W0+Kmax, with Kmax=eV0 (stopping potential) and W0=hu0 (threshold frequency u0).
Step-by-step (primary).
- Work function: W0=hu0=(6.63×10−34)(8×1014)=5.30×10−19J =1.6×10−195.30×10−19≈3.3 eV.
- Maximum KE =eV0=3.3 eV.
- Incident frequency: hν=W0+eV0=(3.3+3.3) eV=6.6 eV=1.056×10−18J. ν=6.63×10−341.056×10−18≈1.6×1015 Hz. …
- AHSEC Higher Secondary (HS) Final Examination 2025Set ANNUAL1 markQ.Fill up the blanks in the following equation: V0 = (h/)ν − (/e)
›Reveal solutionSolution
The equation is Einstein's photoelectric equation rearranged for stopping potential; blanks = e and φ0.
Einstein's photoelectric equation states that the maximum kinetic energy of an emitted photoelectron equals the photon energy minus the work function:
Kmax = hν − φ0
Since the stopping potential V0 is defined by eV0 = Kmax (the retarding potential just able to stop the fastest photoelectron),
eV0 = hν − φ0
Dividing throughout by e:
V0 = (h/e)ν − (φ0/e)
…
- AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL1 markQ.Guess the shape of the curve which shows the variation of V0 with ν in the case of photoelectric emission shown by the relation V0 = (h/e)ν − φ0/e, where the symbols have their usual meaning. OR The de Broglie wavelength of a heavier particle is _______. (Fill up the blank)
›Reveal solutionSolution
The photoelectric stopping-potential equation is linear in ν, so the V₀–ν graph is a straight line; and de Broglie wavelength λ = h/(mv) is inversely proportional to mass, so a heavier particle has a smaller wavelength.
Main question: Einstein's photoelectric equation for the stopping potential is
V0=ehν−eϕ0
This has exactly the form of a straight line, y=mx+c, with y=V0, x=ν, slope m=h/e (a universal positive constant, same for all metals), and intercept c=−ϕ0/e (negative, and metal-dependent through the work function ϕ0). So the graph of V0 against ν is a straight line — it rises with positive slope h/e, crosses the ν-axis at the threshold frequency ν0=ϕ0/h (below which no photoelectrons are emitted), and has a negative V0-intercept of −ϕ0/e.
…
- AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL1 markQ.If the work function of two metals X and Y are 4.17 eV and 8.24×10^-19 J respectively, then for which metal lesser amount of energy will be required to emit an electron?
›Reveal solutionSolution
Converting both work functions to the same unit (eV) shows X (4.17 eV) is smaller than Y (≈5.15 eV), so X needs less energy to emit an electron.
The work function ϕ0 is the minimum energy needed to just eject an electron from a metal surface (with zero kinetic energy). To compare the two given values, convert Y's work function from joules to electron-volts using 1eV=1.6×10−19J:
ϕ0,Y=1.6×10−198.24×10−19=5.15eV
…
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