Photoelectric Effect
Kinetic energy of ejected electrons.
Formula
Variables
How it works
In the photoelectric effect, light arrives as individual photons, each carrying energy hf. A metal can only emit an electron if one photon supplies at least the work function \(\phi\), the minimum energy needed to liberate that electron; any leftover energy becomes the electron’s maximum kinetic energy, so \(K_{max}=hf-\phi\). This is why increasing light intensity alone does not increase electron speed below threshold: intensity changes the number of photons, while frequency changes the energy per photon.
When to use it
Use this formula when a problem involves electrons emitted from a metal surface under illumination and asks for their maximum kinetic energy, stopping potential, threshold frequency, or work function. It is also the right tool for interpreting photoelectric experiments and comparing different light frequencies on the same material.
Worked example, step by step
A sodium surface has a work function of 2.28 eV and is illuminated with light of frequency 8.20 × 10^14 Hz. Step 1: Calculate photon energy: hf = (6.626 × 10^-34 J·s)(8.20 × 10^14 s^-1) = 5.435 × 10^-19 J Step 2: Convert the work function to joules: \phi = 2.28 eV × (1.602 × 10^-19 J/eV) = 3.65 × 10^-19 J Step 3: Subtract to find maximum kinetic energy: K_max = hf - \phi = (5.435 × 10^-19) - (3.65 × 10^-19) = 1.79 × 10^-19 J Step 4: Optional conversion to eV: K_max = (1.79 × 10^-19 J) / (1.602 × 10^-19 J/eV) = 1.12 eV Final answer: K_max = 1.79 × 10^-19 J, or about 1.12 eV
Common mistakes to avoid
- Using wavelength directly without first converting to frequency or photon energy
- Forgetting to convert the work function and photon energy into the same units before subtracting
- Assuming the emitted electrons all have the same kinetic energy instead of only the maximum value
Did You Know?
Einstein won his Nobel Prize for the photoelectric effect, not relativity, cementing that light comes in particles.
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