Photoelectric Effect — Worked Examples

Collection of worked examples for the photoelectric effect, including problems that frequently appear on exams.

Example 1: Finding Work Function and Planck's Constant (EXAM FAVORITE)

[!warning] This problem type appears on almost every exam!

Problem

In a photoelectric effect experiment, a metal surface is illuminated with two different light sources:

Light source 1 2
Frequency $6.0 \times 10^{14}$ Hz $9.0 \times 10^{14}$ Hz
Stopping potential 0.4 V 1.6 V

Calculate: (a) The work function $\phi$ (b) Planck's constant $h$

Solution

Step 1: Write the photoelectric equation

$$eV_s = hf - \phi$$

Where:

  • $e = 1.6 \times 10^{-19}$ C (electron charge)
  • $V_s$ = stopping potential
  • $h$ = Planck's constant
  • $f$ = frequency
  • $\phi$ = work function

Step 2: Set up equations for both data points

For light source 1: $$e(0.4) = h(6.0 \times 10^{14}) - \phi$$ $$0.4e = 6.0 \times 10^{14}h - \phi \quad \text{... (1)}$$

For light source 2: $$e(1.6) = h(9.0 \times 10^{14}) - \phi$$ $$1.6e = 9.0 \times 10^{14}h - \phi \quad \text{... (2)}$$

Step 3: Solve for $h$ by subtracting equations

Subtract (1) from (2): $$(1.6 - 0.4)e = (9.0 - 6.0) \times 10^{14}h$$

$$1.2e = 3.0 \times 10^{14}h$$

$$h = \frac{1.2e}{3.0 \times 10^{14}} = \frac{1.2 \times 1.6 \times 10^{-19}}{3.0 \times 10^{14}}$$

$$h = \frac{1.92 \times 10^{-19}}{3.0 \times 10^{14}} = 6.4 \times 10^{-34} \text{ J·s}$$

[!tip] The accepted value is $h = 6.63 \times 10^{-34}$ J·s. Small differences are due to rounding in the problem data.

Step 4: Solve for $\phi$ using equation (1)

$$0.4e = 6.0 \times 10^{14}h - \phi$$

$$\phi = 6.0 \times 10^{14}h - 0.4e$$

$$\phi = 6.0 \times 10^{14}(6.4 \times 10^{-34}) - 0.4(1.6 \times 10^{-19})$$

$$\phi = 3.84 \times 10^{-19} - 0.64 \times 10^{-19}$$

$$\phi = 3.2 \times 10^{-19} \text{ J}$$

Or in electron volts: $$\phi = \frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0 \text{ eV}$$

Answer

  • Planck's constant: $h = 6.4 \times 10^{-34}$ J·s
  • Work function: $\phi = 3.2 \times 10^{-19}$ J or 2.0 eV

Key Strategy for This Problem Type

When given two (frequency, stopping potential) pairs:

  1. Write the photoelectric equation for each data point
  2. Subtract the equations to eliminate $\phi$
  3. Solve for $h$ first
  4. Substitute back to find $\phi$

Why This Works

The slope of $V_s$ vs $f$ graph gives $h/e$: $$V_s = \frac{h}{e}f - \frac{\phi}{e}$$

This is a linear equation ($y = mx + c$) where:

  • Slope $m = h/e$
  • Y-intercept $c = -\phi/e$

Exam Tips from Lecturer

[!important] Exam Intel The lecturer will give 3 cases for work functions in the photoelectric effect exam question. Be prepared to identify which case applies based on the given data.

Typical 3 Cases:

Case Situation What to Calculate
Case 1 Given $f$ and $V_s$ (2 data points) Find $h$ and $\phi$
Case 2 Given photon energy and $K_{max}$ Find $\phi$ directly: $\phi = E - K_{max}$
Case 3 Given threshold frequency $f_0$ Find $\phi = hf_0$

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