Photoelectric Effect Graphs: Stopping Potential vs. ν, I vs. V | JEE Physics

Why These Graphs Are Crucial for JEE

Graphical analysis of Photoelectric Effect appears in 85% of JEE Main papers and is a favorite in JEE Advanced. Understanding these graphs helps you:

Extract Planck's constant from experimental data
Determine work function of different metals
Understand saturation current and its dependence on intensity
Solve numerical problems quickly using graphical insights

Graph 1 High Yield

Stopping Potential (V₀) vs Frequency (ν) Graph

Graph Visualization: Straight line with positive slope

[V₀ on Y-axis, ν on X-axis - Linear plot starting from negative V₀ at threshold frequency]

📈 Key Features & Interpretation:

Slope Interpretation:

From Einstein's equation: $eV_0 = h\nu - \phi$

$V_0 = \frac{h}{e}\nu - \frac{\phi}{e}$

Slope = $\frac{h}{e}$ (Planck's constant divided by electron charge)

X-intercept Interpretation:

When $V_0 = 0$, $\nu = \nu_0$ (threshold frequency)

$h\nu_0 = \phi$ (Work function)

X-intercept gives threshold frequency

Y-intercept Interpretation:

When $\nu = 0$, $V_0 = -\frac{\phi}{e}$

Y-intercept gives work function (in eV)

🎯 JEE Application Example:

Problem: In a photoelectric experiment, the stopping potential vs frequency graph has slope 4.12 × 10⁻¹⁵ V-s. Find Planck's constant.

Solution: Slope = $\frac{h}{e}$ = 4.12 × 10⁻¹⁵

$h = \text{slope} \times e = (4.12 \times 10^{-15}) \times (1.6 \times 10^{-19})$

$h = 6.59 \times 10^{-34}$ J-s ✓

Graph 2 Important

Photocurrent (I) vs Applied Voltage (V) Graph

Graph Visualization: Saturation curve with sharp cut-off

[I on Y-axis, V on X-axis - Curve rises quickly, then saturates; stopping potential where I=0]

📈 Key Features & Interpretation:

Saturation Current:

• Independent of applied voltage

• Directly proportional to light intensity

• Represents maximum electron emission rate

Stopping Potential (V₀):

• Point where photocurrent becomes zero

• Independent of light intensity

• Depends only on frequency and work function

Effect of Intensity:

• Higher intensity → Higher saturation current

• Same stopping potential for same frequency

• Parallel curves with different saturation levels

Effect of Frequency:

• Higher frequency → More negative stopping potential

• Same saturation current for same intensity

• Curves shift horizontally

🎯 JEE Application Example:

Problem: When light intensity is doubled in photoelectric experiment, what happens to saturation current and stopping potential?

Solution:

• Saturation current doubles (directly proportional to intensity)

• Stopping potential remains unchanged (depends only on frequency)

Advanced Analysis JEE Advanced

Comparing Different Metals & Frequencies

🔬 Multiple Curves Analysis:

Different Metals (Same Frequency):

• Higher work function → More negative Y-intercept in V₀ vs ν graph

• Same slope ($\frac{h}{e}$) for all metals

• Different threshold frequencies

Different Frequencies (Same Metal):

• Higher frequency → More negative stopping potential in I vs V graph

• Same saturation current for same intensity

• Parallel curves shifted horizontally

🎯 JEE Advanced Example:

Problem: Metals A and B have work functions 2eV and 4eV respectively. Compare their V₀ vs ν graphs.

Solution:

• Same slope ($\frac{h}{e}$) for both metals

• Different X-intercepts: ν₀(A) < ν₀(B) since φ(A) < φ(B)

• Different Y-intercepts: V₀ = -φ/e at ν=0

• Metal B's graph is shifted right and down compared to Metal A

🚀 Problem-Solving Strategies

Graph Interpretation Rules:

Slope of V₀ vs ν = h/e (universal constant)
X-intercept = threshold frequency
Saturation current ∝ intensity
Stopping potential depends only on ν

JEE Exam Tips:

Always check units in slope calculations
Remember h = 6.63 × 10⁻³⁴ J-s for verification
Work function in eV = |Y-intercept|
Practice identifying graphs from descriptions

Experimental Variations Available

Includes effect of filtering, different cathode materials, and intensity variations with complete graphical analysis

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. If slope of V₀ vs ν graph is 4.14 × 10⁻¹⁵ V-s, find work function when threshold frequency is 1.2 × 10¹⁵ Hz.

2. Why does saturation current increase with intensity but stopping potential remains constant?

3. Two metals have work functions 2eV and 3eV. Compare their V₀ vs ν graphs.

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