Cracking Photoelectric Effect Problems in 4 Simple Steps
Master the systematic approach to solve any Photoelectric Effect problem in JEE Main and Advanced.
Why This 4-Step Method Works
Photoelectric Effect questions follow predictable patterns in JEE. This systematic approach ensures you never miss key concepts and can solve even the trickiest problems efficiently.
🎯 JEE Relevance
Photoelectric Effect appears in 1-2 questions per JEE paper, carrying 3-4 marks each. Mastering this topic can give you a significant advantage in Modern Physics section.
🚀 The 4-Step Framework
Essential Formulas You Must Know
Einstein's Photoelectric Equation
Where:
$E_k$ = Maximum kinetic energy of photoelectrons
$h$ = Planck's constant ($6.626 \times 10^{-34}$ Js)
$\nu$ = Frequency of incident light
$\nu_0$ = Threshold frequency
$\phi$ = Work function
Key Relationships
Stopping Potential:
$ eV_0 = E_k = h\nu - h\nu_0 $
Work Function:
$ \phi = h\nu_0 = \frac{hc}{\lambda_0} $
Kinetic Energy vs Frequency:
$ E_k = h(\nu - \nu_0) $
Identify Threshold Frequency (ν₀) or Work Function (φ)
Example Problem
A metal surface has work function 2.3 eV. Light of wavelength 400 nm is incident on it. Calculate the maximum kinetic energy of emitted electrons.
Step 1.1: Convert units if necessary
Work function: $\phi = 2.3$ eV
Convert to joules: $\phi = 2.3 \times 1.6 \times 10^{-19} = 3.68 \times 10^{-19}$ J
Step 1.2: Find threshold frequency
Using $\phi = h\nu_0$:
$\nu_0 = \frac{\phi}{h} = \frac{3.68 \times 10^{-19}}{6.626 \times 10^{-34}} = 5.55 \times 10^{14}$ Hz
💡 Key Insights
- Threshold frequency is the minimum frequency needed for photoemission
- If $\nu < \nu_0$, no photoelectrons are emitted regardless of intensity
- Work function is a material property - different for each metal
- Common conversions: 1 eV = $1.6 \times 10^{-19}$ J
Apply Einstein's Photoelectric Equation
Continuing Our Example
Step 2.1: Find frequency of incident light
Wavelength $\lambda = 400$ nm = $400 \times 10^{-9}$ m
$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{400 \times 10^{-9}} = 7.5 \times 10^{14}$ Hz
Step 2.2: Apply Einstein's equation
$E_k = h\nu - h\nu_0 = h(\nu - \nu_0)$
$E_k = 6.626 \times 10^{-34} \times (7.5 - 5.55) \times 10^{14}$
$E_k = 6.626 \times 10^{-34} \times 1.95 \times 10^{14} = 1.29 \times 10^{-19}$ J
In eV: $E_k = \frac{1.29 \times 10^{-19}}{1.6 \times 10^{-19}} = 0.806$ eV
💡 Common Variations
- Sometimes given in terms of wavelength: $E_k = \frac{hc}{\lambda} - \phi$
- If $\nu < \nu_0$, $E_k$ is negative ⇒ no photoemission
- Intensity affects number of photoelectrons, not their energy
- Remember: $hc = 1240$ eV·nm (useful conversion)
Find Stopping Potential (V₀)
Complete the Calculation
Step 3.1: Use stopping potential relation
$eV_0 = E_k$
$V_0 = \frac{E_k}{e} = \frac{1.29 \times 10^{-19}}{1.6 \times 10^{-19}} = 0.806$ V
Final Answers:
- Maximum kinetic energy = 0.806 eV
- Stopping potential = 0.806 V
- Threshold frequency = $5.55 \times 10^{14}$ Hz
💡 Physical Significance
- Stopping potential measures maximum kinetic energy of photoelectrons
- $V_0$ vs $\nu$ graph gives straight line with slope $h/e$
- X-intercept gives threshold frequency
- Y-intercept gives -φ/e
Analyze Graphs and Experimental Data
Key Graphs in Photoelectric Effect
1. $E_k$ vs $\nu$ Graph
Straight line with slope = h, intercept = -φ
Equation: $E_k = h\nu - \phi$
2. $V_0$ vs $\nu$ Graph
Straight line with slope = h/e
Equation: $eV_0 = h\nu - \phi$
3. Photocurrent vs Voltage Graphs
Saturation current depends on intensity, stopping potential depends on frequency
💡 Graph Analysis Tips
- Slope of $V_0$ vs $\nu$ graph = $h/e$ (can be used to find h)
- X-intercept = threshold frequency $\nu_0$
- Changing intensity shifts saturation current, not stopping potential
- Changing frequency shifts stopping potential and saturation current
Complete JEE-Level Problem
JEE Main 2022 Problem
Light of wavelength 248 nm is incident on a cesium surface ($\phi = 1.95$ eV). Calculate:
- Threshold wavelength for cesium
- Maximum kinetic energy of photoelectrons
- Stopping potential
Step 1: Find threshold wavelength
$\phi = 1.95$ eV = $1.95 \times 1.6 \times 10^{-19}$ J
$\lambda_0 = \frac{hc}{\phi} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.95 \times 1.6 \times 10^{-19}} = 6.37 \times 10^{-7}$ m = 637 nm
Step 2: Apply Einstein's equation
Using $hc = 1240$ eV·nm (convenient units):
$E_k = \frac{hc}{\lambda} - \phi = \frac{1240}{248} - 1.95 = 5.0 - 1.95 = 3.05$ eV
Step 3: Find stopping potential
$V_0 = \frac{E_k}{e} = 3.05$ V
⚠️ Common Mistakes to Avoid
Conceptual Errors
- Thinking intensity affects photoelectron energy
- Confusing threshold frequency with threshold wavelength
- Forgetting to convert eV to joules (or vice versa)
- Mixing up slope interpretations in graphs
Calculation Errors
- Using wrong value of Planck's constant
- Not paying attention to units (nm vs m, eV vs J)
- Sign errors in Einstein's equation
- Misreading graph intercepts and slopes
📝 Practice Problems
Problem 1: A metal has work function 4.2 eV. Light of wavelength 200 nm is incident. Will photoemission occur? If yes, find stopping potential.
Problem 2: The stopping potential for light of wavelength 4000 Å is 1.3 V. Find the work function of the metal.
Problem 3: Light of frequencies 1.5ν₀ and 2ν₀ are incident on a metal surface. Find the ratio of maximum kinetic energies of photoelectrons.
📋 4-Step Method Summary
Step-by-Step Checklist
- Find ν₀/φ: Identify threshold frequency or work function
- Einstein's Eq: Apply $E_k = h\nu - \phi$
- Stopping Potential: Use $V_0 = E_k/e$
- Graph Analysis: Interpret slopes and intercepts
Key Constants & Formulas
- $h = 6.626 \times 10^{-34}$ Js
- $hc = 1240$ eV·nm
- 1 eV = $1.6 \times 10^{-19}$ J
- $\phi = h\nu_0 = hc/\lambda_0$
- $E_k = h\nu - \phi = eV_0$
Mastered Photoelectric Effect?
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