Photoelectric Effect: When Light Kicks Out Electrons

The Oppenheimer Connection 🎬

Remember that scene in Oppenheimer where young Einstein revolutionizes physics? The photoelectric effect was his Nobel Prize work - not relativity! This simple experiment where light ejects electrons from metal surfaces destroyed classical physics and birthed quantum mechanics.

“Quantum mechanics is very impressive, but I am convinced God does not play dice.” - Einstein (who ironically started it all)

What is the Photoelectric Effect?

When light of sufficiently high frequency strikes a metal surface, electrons are ejected from the surface. These emitted electrons are called photoelectrons.

Classical Wave Theory’s Epic Fail

According to classical physics:

• Light intensity ↑ → Kinetic energy of electrons ↑
• Any frequency should work if intense enough
• There should be a time lag for electrons to absorb energy

Reality crushed all three predictions! 🎯

Experimental Observations (Lenard, 1902)

The Setup

flowchart LR
    L["Light"] --> C["Cathode
(Metal)"]
    C --> PE["Photoelectrons"]
    C --> PC["Photocurrent (I)"]

Key Observations

1. Threshold Frequency (ν₀)

   • Below ν₀: No emission, regardless of intensity
   • Above ν₀: Instant emission

2. Saturation Current

   • Proportional to light intensity
   • Independent of frequency

3. Stopping Potential (V₀)

   • Depends on frequency, NOT intensity
   • V₀ ∝ ν (linear relationship)

4. Instantaneous Emission

   • No time lag (~10⁻⁹ s)

Einstein’s Quantum Explanation (1905)

The Revolutionary Idea

Light consists of discrete packets of energy called photons:

where:

• h = 6.626 × 10⁻³⁴ J·s (Planck’s constant)
• ν = frequency
• c = 3 × 10⁸ m/s
• λ = wavelength

Einstein’s Photoelectric Equation

:::box Master Formula

or equivalently:

where:

• K_max = Maximum kinetic energy of photoelectrons
• φ = Work function (minimum energy to remove electron)
• hν = Energy of incident photon :::

In Terms of Wavelength

Work Function

where ν₀ is the threshold frequency.

Stopping Potential

The stopping potential (V₀) is the minimum negative potential applied to anode that stops the most energetic photoelectrons.

:::box Key Relations

This is a straight line equation: y = mx + c

• Slope = h/e (universal constant)
• y-intercept = -φ/e :::

Interactive Demo: Visualize Photoelectric Effect

See how light frequency and intensity affect electron emission in real-time.

The Million-Dollar Graph

V₀ (V)
  ↑
  |     /
  |    /  Slope = h/e = 4.14 × 10⁻¹⁵ V·s
  |   /
  |  /
  | /
──┼───────→ ν (Hz)
  |/ν₀

Intercept on ν-axis = ν₀ = φ/h
Intercept on V₀-axis = -φ/e

Interactive Demo: Photoelectric Simulator

// Adjust frequency and intensity
const PhotoelectricDemo = () => {
  const [frequency, setFrequency] = useState(5e14); // Hz
  const [intensity, setIntensity] = useState(50); // arbitrary units

  // Sodium work function
  const workFunction = 2.3; // eV
  const h = 4.14e-15; // eV·s

  const photonEnergy = h * frequency;
  const kMax = photonEnergy - workFunction;
  const emissionOccurs = photonEnergy > workFunction;

  return (
    <div>
      <h3>Photoelectric Effect Simulator</h3>
      <p>Work Function (Sodium): {workFunction} eV</p>

      <label>Frequency: {frequency.toExponential(2)} Hz</label>
      <input
        type="range"
        min="1e14"
        max="1e15"
        value={frequency}
        onChange={(e) => setFrequency(Number(e.target.value))}
      />

      <label>Intensity: {intensity}%</label>
      <input
        type="range"
        min="0"
        max="100"
        value={intensity}
        onChange={(e) => setIntensity(Number(e.target.value))}
      />

      <div style={{backgroundColor: emissionOccurs ? 'green' : 'red'}}>
        {emissionOccurs ? (
          <>
            <p>✓ Photoelectron Emission!</p>
            <p>Photon Energy: {photonEnergy.toFixed(2)} eV</p>
            <p>K_max: {kMax.toFixed(2)} eV</p>
            <p>Number of electrons ∝ {intensity}</p>
          </>
        ) : (
          <>
            <p>✗ No Emission</p>
            <p>Photon Energy: {photonEnergy.toFixed(2)} eV</p>
            <p>Need: {workFunction} eV minimum</p>
          </>
        )}
      </div>
    </div>
  );
};

Memory Tricks 🧠

“STOP” for Stopping Potential

• Stopping potential depends on ν
• Threshold frequency exists
• One photon → one electron
• Photocurrent depends on intensity

Work Function Ladder

Low φ  → Easy to liberate electrons → Alkali metals
High φ → Hard to liberate electrons → Transition metals

Cs < Na < K < Al < Cu < Zn < Ag < Pt
↓                                    ↑
Easy                              Hard

Einstein’s Energy Budget

Photon brings hν
    ↓
Pay φ (entrance fee)
    ↓
Leftover = K_max (kinetic energy)

Common Mistakes ⚠️

❌ Mistake 1: Intensity determines emission

Wrong: “Shine brighter light to eject electrons from any metal” Right: Frequency must exceed threshold; intensity only affects number of electrons

❌ Mistake 2: All electrons have K_max

Wrong: “All photoelectrons have the same kinetic energy” Right: Electrons have 0 ≤ K ≤ K_max depending on depth in metal

❌ Mistake 3: Negative stopping potential

Wrong: “V₀ = -2 V means apply +2V to stop electrons” Right: Apply -2V to cathode (or +2V to anode relative to cathode)

❌ Mistake 4: Wavelength confusion

Wrong: K_max = hλ - φ Right: K_max = hc/λ - φ (inverse relationship!)

❌ Mistake 5: Time lag exists

Wrong: “Electrons need time to accumulate energy” Right: Emission is instantaneous (~10⁻⁹ s)

Derivations You Must Know

1. Relation between ν₀ and λ₀

At threshold: K_max = 0

2. Ratio of Stopping Potentials

For two frequencies ν₁ and ν₂:

Special case: If ν₂ = ν₀ (threshold), then V₀₂ = 0

3. Change in Stopping Potential

Important Constants & Values

Work Functions of Common Metals

Problem-Solving Strategy

Step 1: Identify what’s given

• Frequency/wavelength
• Intensity
• Metal (work function)
• Stopping potential

Step 2: Choose the right equation

• For K_max: Use hν - φ
• For V₀: Use eV₀ = K_max
• For threshold: Use hν₀ = φ

Step 3: Unit consistency

• Convert nm → m (×10⁻⁹)
• Convert eV ↔ J (×1.6×10⁻¹⁹)
• Use hc = 1240 eV·nm for quick calculations

Practice Problems

Level 1: JEE Main Basics

Q1. Light of wavelength 400 nm falls on a metal surface with work function 2.0 eV. Calculate the maximum kinetic energy of photoelectrons.

Solution:

Given: λ = 400 nm, φ = 2.0 eV
Using K_max = hc/λ - φ

K_max = 1240 eV·nm / 400 nm - 2.0 eV
K_max = 3.1 - 2.0 = 1.1 eV

Q2. The stopping potential for photoelectrons from a metal surface is 3.2 V when light of wavelength 300 nm is used. Find the work function.

Solution:

Given: V₀ = 3.2 V, λ = 300 nm
eV₀ = hc/λ - φ

φ = hc/λ - eV₀
φ = 1240/300 - 3.2
φ = 4.13 - 3.2 = 0.93 eV

Q3. If the intensity of light is doubled, what happens to: (a) Stopping potential (b) Photocurrent (c) Maximum kinetic energy

Answer:

(a) No change (depends only on frequency)
(b) Doubles (more photons → more electrons)
(c) No change (depends only on frequency)

Level 2: JEE Main/Advanced

Q4. A metal has a threshold wavelength of 500 nm. When light of wavelength 400 nm falls on it, find: (a) Work function (b) Maximum kinetic energy (c) Stopping potential

Solution:

(a) φ = hc/λ₀ = 1240/500 = 2.48 eV

(b) K_max = hc/λ - φ = 1240/400 - 2.48
    K_max = 3.1 - 2.48 = 0.62 eV

(c) V₀ = K_max/e = 0.62 V

Q5. The stopping potential for sodium (φ = 2.3 eV) is 1.5 V. If the same light falls on potassium (φ = 2.3 eV), what is the stopping potential?

Solution:

For sodium: eV₀₁ = hν - φ₁
1.5 = hν - 2.3
hν = 3.8 eV

For potassium: eV₀₂ = hν - φ₂
eV₀₂ = 3.8 - 2.3 = 1.5 V

Answer: Same! (both have same work function)

Q6. A graph of stopping potential vs frequency has a slope of 4.12 × 10⁻¹⁵ V·s. What is the value of Planck’s constant?

Solution:

Slope = h/e
h = slope × e
h = 4.12 × 10⁻¹⁵ × 1.6 × 10⁻¹⁹
h = 6.59 × 10⁻³⁴ J·s

(Close to actual value 6.626 × 10⁻³⁴ J·s)

Level 3: JEE Advanced

Q7. A monochromatic light source of power 5 mW emits 8 × 10¹⁵ photons per second. This light ejects photoelectrons from a metal surface with stopping potential 2.0 V. The efficiency of photoelectron emission is 0.1%. Calculate: (a) Wavelength of light (b) Work function of metal (c) Photocurrent if all emitted electrons reach the anode

Solution:

(a) Power = Number of photons × Energy per photon
    5 × 10⁻³ = 8 × 10¹⁵ × hc/λ
    5 × 10⁻³ = 8 × 10¹⁵ × (6.626 × 10⁻³⁴ × 3 × 10⁸)/λ
    λ = 3.18 × 10⁻⁷ m = 318 nm

(b) eV₀ = hc/λ - φ
    2.0 = 1240/318 - φ
    φ = 3.9 - 2.0 = 1.9 eV

(c) Electrons emitted = 0.1% × 8 × 10¹⁵
                      = 8 × 10¹² electrons/s
    I = ne = 8 × 10¹² × 1.6 × 10⁻¹⁹
    I = 1.28 × 10⁻⁶ A = 1.28 μA

Q8. Two metallic surfaces A and B have work functions 3 eV and 4 eV respectively. Light of wavelength 300 nm falls on both. Find the ratio of maximum velocities of photoelectrons from A and B.

Solution:

For A: K_A = hc/λ - φ_A = 1240/300 - 3 = 1.13 eV
For B: K_B = hc/λ - φ_B = 1240/300 - 4 = 0.13 eV

½mv_A² = K_A and ½mv_B² = K_B

v_A/v_B = √(K_A/K_B) = √(1.13/0.13) = √8.7 = 2.95

Q9. A photon of energy 5 eV ejects a photoelectron from a metal surface. The electron makes a head-on collision with a hydrogen atom in ground state. Find the minimum work function of metal for which the hydrogen atom will be excited to the first excited state. (Ionization energy of H = 13.6 eV)

Solution:

For H: E₁ = -13.6 eV, E₂ = -13.6/4 = -3.4 eV
ΔE = E₂ - E₁ = -3.4 - (-13.6) = 10.2 eV

For excitation, electron needs KE ≥ 10.2 eV

K_max = hν - φ
10.2 = 5 - φ_min
φ_min = -5.2 eV

This is negative! Impossible.

Actually, photon energy must be increased:
φ_max = hν - K_min
φ_max = 5 - 10.2 = -5.2 eV

Correct interpretation: Need hν ≥ 10.2 + φ
For given hν = 5 eV, impossible regardless of φ.

For φ_min: Set K_max = 10.2 eV
φ_min = 5 - 10.2 = -5.2 eV (unphysical)

Answer: Impossible with 5 eV photon. Need at least 10.2 eV photon.

Cross-Links to Related Topics

• de Broglie Hypothesis - Matter waves and wave-particle duality
• Davisson-Germer Experiment - Experimental proof of matter waves
• Bohr Model - Quantization of energy (another quantum success)
• Hydrogen Spectrum - Photon emission and absorption

Quick Revision Checklist ✓

• Einstein’s photoelectric equation: K_max = hν - φ
• Stopping potential: eV₀ = K_max
• Work function: φ = hν₀ = hc/λ₀
• Graph of V₀ vs ν (slope = h/e)
• Four experimental observations
• One photon ejects one electron (intensity → number)
• No time lag, instantaneous emission
• Frequency determines K_max, not intensity
• hc = 1240 eV·nm (golden constant)
• Work functions of common metals

Final Tips for JEE

1. Always check units: nm vs m, eV vs J
2. Use hc = 1240 eV·nm: Saves calculation time
3. Draw the V₀-ν graph: Visual memory helps
4. Remember: Frequency = energy, Intensity = number
5. Practice threshold problems: Very common in JEE
6. Know work functions: At least 3-4 metals

Last updated: April 15, 2025 Next: de Broglie Hypothesis - Wave Nature of Matter