Multi-Chromatic Photoelectric Effect Calculations: Accurate calculation of maximum kinetic energy ($K_{\max}$) and stopping potential when a metallic surface is simultaneously exposed to multiple wavelengths of light.

Multi-Chromatic Photoelectric Effect Calculations

📘 Complete Theory

When a metal surface is illuminated simultaneously by light of different wavelengths, photoelectrons may be emitted due to more than one radiation. However, the maximum kinetic energy of the emitted photoelectrons is determined by the radiation having the highest frequency, i.e. the shortest wavelength, among the radiations capable of producing photoemission.

λ₁, λ₂, λ₃ → Incident Light

Metal Surface

Fastest electron → Kmax

hf = φ + K_max

Therefore, for a single wavelength:

K_max = hf − φ

Since:

f = c/λ

we get:

K_max = hc/λ − φ

Stopping Potential

eV_s = K_max

Therefore:

V_s = (hf − φ)/e

For Multiple Wavelengths

Suppose wavelengths λ₁, λ₂, λ₃ are incident simultaneously. First determine which wavelengths can produce photoelectric emission.

Important Rule:

1. Smaller λ → larger frequency.
2. Larger frequency → larger photon energy.
3. Larger photon energy → larger K_max.
4. Therefore the shortest wavelength which satisfies λ < λ₀ gives the maximum kinetic energy.
5. If λ ≥ λ₀, that radiation cannot eject photoelectrons.

Threshold Wavelength

φ = hc/λ₀

Hence:

λ₀ = hc/φ

Maximum Kinetic Energy

K_max = hc/λ_min,active − φ

Stopping Potential

V_s,max = [hc/λ_min,active − φ]/e

🧮 Fully Solved Example

Question:

A metal has work function 2 eV and is simultaneously illuminated by 400 nm, 500 nm and 700 nm light. Find the maximum kinetic energy and maximum stopping potential.

Step 1: Find photon energies.

Using:
E(eV) ≈ 1240/λ(nm)

For 400 nm:
E = 1240/400 = 3.10 eV

For 500 nm:
E = 1240/500 = 2.48 eV

For 700 nm:
E = 1240/700 ≈ 1.77 eV

Work function = 2 eV.

Therefore 700 nm cannot eject electrons because:
1.77 < 2 eV.

400 nm and 500 nm can eject electrons.

The highest photon energy is from 400 nm.

Therefore:
K_max = 3.10 − 2.00
K_max = 1.10 eV

Since:
eV_s = K_max
V_s = 1.10 V

⚠️ Important Cases

Case 1: All wavelengths are above threshold.
No photoelectrons are emitted.

Case 2: Only one wavelength is below threshold.
That wavelength determines K_max.

Case 3: Several wavelengths are below threshold.
The shortest wavelength determines K_max.

Case 4: Equal intensity but different wavelengths.
Intensity does not determine K_max; frequency determines it.

Case 5: Increasing intensity of one wavelength.
It can increase photoelectric current, but not K_max if its frequency remains unchanged.

📋 Formula Sheet

Quantity	Formula
Photon Energy	E = hf = hc/λ
Einstein Equation	hf = φ + Kmax
Maximum KE	Kmax = hf − φ
Wavelength Form	Kmax = hc/λ − φ
Stopping Potential	eVs = Kmax
Threshold Frequency	f0 = φ/h
Threshold Wavelength	λ0 = hc/φ
Useful Numerical Form	E(eV) = 1240/λ(nm)

🎯 30 MCQs — JEE / NEET / Competitive

⚡ 30 Assertion–Reason Questions

🧮 30 Numerical Problems with Solutions

⭐ 1 Mark Questions — 30

⭐ 2 Marks Questions — 30

⭐ 3 Marks Questions — 30

⭐ 4 Marks Questions — 30

⭐ 5 Marks Questions — 30

⭐ 6 Marks Questions — 30

🚀 Final Revision

K_max = hc/λ − φ

eV_s = K_max

λ₀ = hc/φ

Golden Rule:

For simultaneous illumination by multiple wavelengths, first eliminate wavelengths that cannot cause photoemission. Among the remaining wavelengths, choose the shortest wavelength. That wavelength produces the largest photon energy and therefore the largest K_max and stopping potential.