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Einstein Photoelectric Equation

🎓 Class 12 Physics CBSE Theory Ch 11 – Dual Nature of Radiation and Matter ⏱ ~14 min
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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:

  1. Brighter light should give photoelectrons of higher kinetic energy — more energy soaks into each electron.
  2. The effect should occur at any frequency, provided the beam is intense enough.
  3. 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).

The mismatch: The photoelectric effect simply does not behave as a continuous wave should. Something deeply non-classical is happening at the metal surface.

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:

\(K_\text{max} = \tfrac{1}{2}m v^{2}_\text{max} = h\nu - \phi_0\)

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 \(K_\text{max}\).
One quantum, one electron: The probability of an electron absorbing two photons at once is vanishingly small at ordinary light levels. So the energy budget is set by a single photon — which is why intensity (number of photons) cannot make up for low frequency (insufficient energy per photon).

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.
ν V₀ 0 ν0A Metal A (Cs, low φ₀) ν0B Metal B (Cu, high φ₀) slope = h/e (same for both)
Fig 11.6: Stopping-potential–frequency lines for two metals. Same slope h/e; different x-intercepts give the threshold frequencies, different y-intercepts give −φ₀/e.

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).

Recap — Einstein vs Wave Theory:
ObservationWave theory predictsEinstein's quantum view
Threshold frequencyNone — any \(\nu\) should work\(\nu_0 = \phi_0/h\) — natural
Kmax vs intensityIncreases with intensityIndependent of intensity
Kmax vs frequencyShould be independentLinear: \(K_\text{max}=h\nu-\phi_0\)
Time-lagDetectable for low intensityInstantaneous (one-photon)
Activity 11.2 — Find h/e from a Photocell Plot

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)5.56.57.58.59.5
Stopping potential V₀ (V)0.150.550.951.361.77
Predict: What slope (in V·s) do you expect, and what y-intercept?
Slope from the first and last points: \((1.77-0.15)/(9.5-5.5)\times10^{-14} = 1.62/4 \times 10^{-14} = 0.405\times10^{-14}\) V·s.
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.

10.0
Photon E
4.14 eV
Kmax
2.00 eV
V₀
2.00 V
hν (photon) φ₀ (work) Kmax EMISSION

Worked Examples

Example 1 — Stopping potential of caesium

Light of wavelength 400 nm strikes a caesium surface (φ₀ = 2.14 eV). Find the maximum kinetic energy of photoelectrons and the stopping potential.

Photon energy: \(E = hc/\lambda = (1240\,\text{nm·eV})/400\,\text{nm} = 3.10\) eV.
\(K_\text{max} = E - \phi_0 = 3.10 - 2.14 = 0.96\) eV.
Stopping potential: \(V_0 = K_\text{max}/e = 0.96\) V.
Example 2 — Slope of V₀ vs ν line

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.

\(h = (\text{slope})\times e = 4.12\times10^{-15}\times1.602\times10^{-19} = 6.60\times10^{-34}\) J·s — Planck's constant!
\(\phi_0 = h\nu_0 = 6.60\times10^{-34}\times5\times10^{14} = 3.30\times10^{-19}\) J \(= 2.06\) eV.
Example 3 — Maximum speed of photoelectrons

Sodium (φ₀ = 2.75 eV) is illuminated by light of wavelength 300 nm. Find the maximum speed of the photoelectrons.

\(E = 1240/300 = 4.13\) eV. \(K_\text{max} = 4.13 - 2.75 = 1.38\) eV \(= 2.21\times10^{-19}\) J.
\(v_\text{max} = \sqrt{2K/m} = \sqrt{2\times2.21\times10^{-19}/9.11\times10^{-31}} = 6.97\times10^{5}\) m/s — about 0.23% of c.
Example 4 — Identifying the metal

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.

\(E = h\nu = 6.626\times10^{-34}\times8\times10^{14} = 5.30\times10^{-19}\) J \(= 3.31\) eV.
\(\phi_0 = E - eV_0 = 3.31 - 0.60 = 2.71\) 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:

  • (a) \(h\nu\)
  • (b) \(h\nu + \phi_0\)
  • (c) \(h\nu - \phi_0\)
  • (d) \(\phi_0 - h\nu\)
(c) \(K_\text{max} = h\nu - \phi_0\). The photon energy minus the work function appears as kinetic energy.

Q2. The slope of the V₀-vs-ν graph for any metal equals:

  • (a) \(\phi_0\)
  • (b) \(h/e\)
  • (c) \(\nu_0\)
  • (d) \(e/h\)
(b) The slope is the universal constant \(h/e\) — independent of the metal.

Q3. (Short Answer) Why is the photoelectric effect impossible to explain using a wave picture of light?

A wave delivers energy continuously, so its predictions are: Kmax grows with intensity, no threshold frequency, and a measurable build-up time. None of these match observation.

Q4. (Fill in the blank) Below the ______ frequency, no photoelectric emission occurs however bright the incident light.

Threshold (\(\nu_0 = \phi_0/h\)).

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?

\(\phi_0 = hc/\lambda_0 = 1240/540 = 2.30\) eV. Photon energy = 1240/270 = 4.59 eV. \(eV_0 = 4.59 - 2.30 = 2.29\) eV → V₀ ≈ 2.29 V.

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.

(A) Both correct. The universal slope is the strongest verification of Einstein's quantum picture.

Assertion: Doubling the intensity of incident light doubles the maximum kinetic energy of photoelectrons.

Reason: Intensity is the energy delivered per second per area.

(D) Assertion is false (KE depends only on frequency). The reason is true but does not save the assertion.

Assertion: Two metals exposed to the same UV light may show different stopping potentials.

Reason: They have different work functions.

(A) Both correct and the reason explains the assertion: \(eV_0 = h\nu - \phi_0\) gives a smaller V₀ for the metal with larger \(\phi_0\).

Frequently Asked Questions - Einstein Photoelectric Equation

What is the main concept covered in Einstein Photoelectric Equation?
In NCERT Class 12 Physics Chapter 11 (Dual Nature of Radiation and Matter), "Einstein Photoelectric Equation" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Einstein Photoelectric Equation useful in real-life applications?
Real-life applications of "Einstein Photoelectric Equation" from NCERT Class 12 Physics Chapter 11 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Einstein Photoelectric Equation?
Key formulas in "Einstein Photoelectric Equation" (NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 11?
NCERT Class 12 Physics Chapter 11 (Dual Nature of Radiation and Matter) is structured so each part builds on the previous one. "Einstein Photoelectric Equation" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Einstein Photoelectric Equation?
CBSE board questions from "Einstein Photoelectric Equation" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Einstein Photoelectric Equation" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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