This MCQ module is based on: Einstein Photoelectric Equation
Einstein Photoelectric Equation
This assessment will be based on: Einstein Photoelectric Equation
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Einstein Photoelectric Equation
11.5 Wave Theory Cannot Explain the Photoelectric Effect
If light were a continuous electromagnetic wave (as Maxwell's theory says), the energy delivered to the metal would scale with intensity. Three predictions follow:
- Brighter light should give photoelectrons of higher kinetic energy — more energy soaks into each electron.
- The effect should occur at any frequency, provided the beam is intense enough.
- For a faint beam, the electron should accumulate energy slowly; a measurable time-lag of seconds, even minutes, should appear before emission begins.
Every one of these predictions fails when measured. The maximum kinetic energy depends on frequency only; below \(\nu_0\) no current flows however bright the beam; and the emission, even at the lowest detectable intensities, is essentially instantaneous (\(<10^{-9}\) s).
11.6 Einstein's Photoelectric Equation — Energy Quantum of Radiation
In 1905, working from Planck's 1900 hypothesis that radiation is emitted in discrete packets of energy \(h\nu\), Albert Einstein proposed a radically simple resolution: light itself is granular. A monochromatic beam of frequency \(\nu\) is a stream of energy quanta, each of energy \(h\nu\). When such a quantum is absorbed by an electron in the metal, the entire energy \(h\nu\) is transferred in one indivisible event.
Of this energy, the electron must spend at least \(\phi_0\) (the work function) to escape the surface. Whatever remains shows up as kinetic energy of the freed photoelectron. The most loosely bound electrons (those at the very surface) lose only \(\phi_0\); deeper-lying electrons lose more, so they emerge with smaller kinetic energy. Hence:
Einstein's photoelectric equation (Eq 11.2)
Three immediate consequences follow naturally:
- Threshold frequency: Photoelectrons can be liberated only if \(h\nu \geq \phi_0\), i.e. \(\nu \geq \nu_0\) where \[\nu_0 = \frac{\phi_0}{h}\]
- Linear KE–frequency relation: \(K_\text{max}\) varies linearly with \(\nu\); the slope is the same for every metal — the universal constant \(h\).
- Intensity ↔ photon number: A more intense beam carries more photons per second but each still has energy \(h\nu\). So intensity sets the saturation current, not \(K_\text{max}\).
11.7 Stopping Potential and Work Function
Multiplying the equation by 1 and using \(K_\text{max} = eV_0\) (where \(V_0\) is the stopping potential measured in the experiment):
\[eV_0 = h\nu - \phi_0 \quad\Longrightarrow\quad V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}\]This is exactly the linear graph observed by Hallwachs and Lenard. A plot of \(V_0\) versus \(\nu\) gives:
- A straight line of slope \(h/e\) — the same for every metal.
- An x-intercept at \(\nu_0 = \phi_0/h\) — different for each metal.
- A y-intercept at \(-\phi_0/e\) — directly giving the work function.
11.8 Millikan's Verification (1916)
Robert A. Millikan, originally sceptical of Einstein's quantum hypothesis, performed precise measurements of \(V_0\) versus \(\nu\) for several alkali metals. His results were a perfect straight line with the predicted slope \(h/e\). Once he inserted the known electronic charge \(e\) (which he himself had measured in the famous oil-drop experiment), the value of Planck's constant emerged:
\[h = 6.626\times10^{-34}\,\text{J·s}\]This agreement, both quantitative and structural, settled the case in favour of Einstein's quantum picture and led to the 1921 Nobel Prize for the photoelectric law (and the 1923 Nobel Prize for Millikan).
| Observation | Wave theory predicts | Einstein's quantum view |
|---|---|---|
| Threshold frequency | None — any \(\nu\) should work | \(\nu_0 = \phi_0/h\) — natural |
| Kmax vs intensity | Increases with intensity | Independent of intensity |
| Kmax vs frequency | Should be independent | Linear: \(K_\text{max}=h\nu-\phi_0\) |
| Time-lag | Detectable for low intensity | Instantaneous (one-photon) |
Given the table of stopping potentials from a sodium photocell experiment, plot V₀ vs ν and read off h/e and the work function.
| Frequency \(\nu\) (1014 Hz) | 5.5 | 6.5 | 7.5 | 8.5 | 9.5 |
|---|---|---|---|---|---|
| Stopping potential V₀ (V) | 0.15 | 0.55 | 0.95 | 1.36 | 1.77 |
This equals \(h/e\), so \(h = 0.405\times10^{-14}\times1.6\times10^{-19} = 6.48\times10^{-34}\) J·s — within 2% of the textbook value!
Threshold frequency from x-intercept: extrapolating, V₀=0 at \(\nu_0\approx5.13\times10^{14}\) Hz, giving \(\phi_0 = h\nu_0 = 2.12\) eV — matches sodium (textbook \(\phi_0=2.75\) eV; our data are stylised for the activity).
Interactive — Einstein's Equation Explorer
Pick a metal and slide the photon frequency. Watch \(K_\text{max}\) and the stopping potential update in real time. Notice that no photoelectrons emerge below the threshold frequency.
4.14 eV
2.00 eV
2.00 V
Worked Examples
Light of wavelength 400 nm strikes a caesium surface (φ₀ = 2.14 eV). Find the maximum kinetic energy of photoelectrons and the stopping potential.
\(K_\text{max} = E - \phi_0 = 3.10 - 2.14 = 0.96\) eV.
Stopping potential: \(V_0 = K_\text{max}/e = 0.96\) V.
An experiment yields a straight-line plot of V₀ vs ν with slope \(4.12\times10^{-15}\) V·s and x-intercept at \(5\times10^{14}\) Hz. Compute \(h\) and the metal's work function.
\(\phi_0 = h\nu_0 = 6.60\times10^{-34}\times5\times10^{14} = 3.30\times10^{-19}\) J \(= 2.06\) eV.
Sodium (φ₀ = 2.75 eV) is illuminated by light of wavelength 300 nm. Find the maximum speed of the photoelectrons.
\(v_\text{max} = \sqrt{2K/m} = \sqrt{2\times2.21\times10^{-19}/9.11\times10^{-31}} = 6.97\times10^{5}\) m/s — about 0.23% of c.
Light of frequency \(8\times10^{14}\) Hz produces photoelectrons of stopping potential 0.6 V. Find the work function and identify the metal from Table 11.1.
\(\phi_0 = E - eV_0 = 3.31 - 0.60 = 2.71\) eV — matches sodium (φ₀ = 2.75 eV).
Competency-Based Questions
Q1. According to Einstein's photoelectric equation, the maximum kinetic energy of a photoelectron is:
Q2. The slope of the V₀-vs-ν graph for any metal equals:
Q3. (Short Answer) Why is the photoelectric effect impossible to explain using a wave picture of light?
Q4. (Fill in the blank) Below the ______ frequency, no photoelectric emission occurs however bright the incident light.
Q5. (HOT) The threshold wavelength for a metal is 540 nm. What is the stopping potential when light of wavelength 270 nm is incident on it?
Assertion–Reason Questions
Options: (A) Both true, R correct explanation of A. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: The slope of the stopping-potential vs frequency graph is the same for all metals.
Reason: This slope equals h/e, where h is Planck's constant and e is the electronic charge.
Assertion: Doubling the intensity of incident light doubles the maximum kinetic energy of photoelectrons.
Reason: Intensity is the energy delivered per second per area.
Assertion: Two metals exposed to the same UV light may show different stopping potentials.
Reason: They have different work functions.
Frequently Asked Questions - Einstein Photoelectric Equation
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Board exam sample papers
Physics — CBSE Class XII Sample Paper 1 (2025-26)
Section A · Section B · Section C · Section D · Section E