Photoelectric Effect: Common Mistakes and Fixes (2)

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Tags Photoelectric Effect

Question

Light of frequency ν=8×1014 Hz\nu = 8 \times 10^{14}\,\text{Hz} falls on a metal of work function ϕ=2 eV\phi = 2\,\text{eV}. Many students multiply intensity to find the maximum kinetic energy of emitted electrons. Why is that wrong, and what is the correct Kmax⁡K_{\max}? Take h=6.63×10−34 Jsh = 6.63\times 10^{-34}\,\text{Js} and 1 eV=1.6×10−19 J1\,\text{eV} = 1.6\times 10^{-19}\,\text{J}.

Solution — Step by Step

Each photon carries energy hνh\nu. If this exceeds the work function, the excess becomes the electron’s kinetic energy:

Kmax⁡=hν−ϕK_{\max} = h\nu - \phi

Intensity affects the number of photoelectrons, not their energy.

hν=6.63×10−34×8×1014=5.30×10−19 J=3.31 eVh\nu = 6.63\times 10^{-34} \times 8\times 10^{14} = 5.30\times 10^{-19}\,\text{J} = 3.31\,\text{eV}

Kmax⁡=3.31−2.00=1.31 eVK_{\max} = 3.31 - 2.00 = 1.31\,\text{eV}

In SI units: Kmax⁡≈2.10×10−19 JK_{\max} \approx 2.10\times 10^{-19}\,\text{J}.

The maximum kinetic energy is Kmax⁡≈1.31 eVK_{\max} \approx 1.31\,\text{eV}.

Why This Works

Einstein’s quantum picture says light interacts with electrons one photon at a time. A photon either has enough energy to liberate an electron or it doesn’t. Doubling the intensity doubles the photon count, so doubles the photocurrent — but each individual photon still has the same energy hνh\nu, so Kmax⁡K_{\max} is unchanged.

This was the experimental fact that classical wave theory couldn’t explain. Increasing wave amplitude (intensity) should have given electrons more energy, but it didn’t. That’s how photons were born.

Alternative Method

Use SI units throughout. ϕ=2×1.6×10−19=3.2×10−19 J\phi = 2 \times 1.6\times 10^{-19} = 3.2\times 10^{-19}\,\text{J}.

Kmax⁡=5.30×10−19−3.20×10−19=2.10×10−19 JK_{\max} = 5.30\times 10^{-19} - 3.20\times 10^{-19} = 2.10\times 10^{-19}\,\text{J}

Convert: Kmax⁡/(1.6×10−19)=1.31 eVK_{\max}/(1.6\times 10^{-19}) = 1.31\,\text{eV}. Same answer.

The classic three confusions:

  1. Intensity changes Kmax⁡K_{\max} — wrong, it changes photocurrent.
  2. Frequency below threshold still emits some electrons — wrong, no emission at all.
  3. Work function depends on the incident light — wrong, it’s a property of the metal.

Common Mistake

Plugging intensity into the energy equation. The fix is to memorise the slogan: frequency controls energy, intensity controls count. JEE Main 2023 had a four-option trap with Kmax⁡K_{\max} values that included the wrong intensity-multiplied answer.

Want to master this topic?

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