Preview

Photoelectric Effect

Discover how light frequency — not intensity — ejects electrons

1905. Physics has an embarrassing problem: shine light on a metal and electrons jump out — but only if the light is the right color. Crank a red lamp to blinding brightness: nothing. A faint blue glow: electrons flow. Einstein's answer sounds insane — light arrives in packets, each priced strictly by color, and an electron is a turnstile that accepts exactly one coin at a time. You are about to replay the experiment that broke classical physics.

What you'll be able to do

  • Explain photoelectron ejection as a single-photon transaction: E_photon = hf versus the metal's work function φ
  • Predict independently how changing frequency versus changing intensity affects the photocurrent and the maximum kinetic energy
  • Apply KE_max = hf − φ and eV_stop = KE_max to compute thresholds, kinetic energies, and stopping voltages

Formulas

E=hf(h=6.626×1034Js)E = hf \quad (h = 6.626 \times 10^{-34}\,\text{J}\cdot\text{s})
Photon Energy
KEmax=hfφKE_{max} = hf - \varphi
Einstein Photoelectric Equation
eVstop=KEmaxeV_{stop} = KE_{max}
Stopping Voltage relation
fthreshold=φhf_{threshold} = \frac{\varphi}{h}
Threshold Frequency

Make a prediction

In the default setup — light at 7.0×10¹⁴ Hz, intensity 5, work function φ = 2.3 eV — the badge reads Electrons Ejecting and the panel shows KE_max = 0.595 eV. You drop the frequency to 4.5×10¹⁴ Hz and crank intensity to the maximum 10 — double the photons per second. What happens?

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Your prediction

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Answer: Emission stops dead. Each 4.5×10¹⁴ Hz photon carries only 1.86 eV — short of the 2.3 eV exit fee — and doubling intensity only doubles the photon count, never the energy per photon

Intensity counts photons per second; frequency prices each one. A photon at 4.5×10¹⁴ Hz carries E = hf = 1.86 eV, below the 2.3 eV work function, so every single photon arrives short of the exit fee — no accumulation, no delay, no matter how many arrive per second; the badge flips to Below Threshold — No Emission and KE reads 0 (no emission). A photon at 7.0×10¹⁴ Hz carries 2.90 eV: it pays the fee with 0.595 eV left over as kinetic energy, even when photons arrive one at a time at intensity 1. That decoupling — frequency for energy, intensity for count — is exactly what the wave picture could not explain, and it won Einstein the 1921 Nobel Prize.

Quiz (0/3)

Why does increasing light intensity below the threshold frequency produce NO photoelectrons?

Sodium has a work function φ = 2.3 eV (this sim's sodium preset). What is the minimum threshold frequency?

Light of f = 5 × 10¹⁴ Hz shines on a surface with work function φ = 4.5 eV (this sim's second preset). Are photoelectrons emitted?

You can now

  • Explain photoelectron ejection as a single-photon transaction: E_photon = hf versus the metal's work function φ
  • Predict independently how changing frequency versus changing intensity affects the photocurrent and the maximum kinetic energy
  • Apply KE_max = hf − φ and eV_stop = KE_max to compute thresholds, kinetic energies, and stopping voltages

Step-by-step

  1. Increase light frequency until the electron counter starts registering — that is the threshold frequency.
  2. Note that raising intensity below threshold never produces electrons.
  3. Use the work-function presets (φ = 2.3, 4.5, 4.7 eV) or drag the φ slider to change the metal surface, and watch the threshold frequency move with it.

Key formulas

  • E=hf(h=6.626×1034Js)E = hf \quad (h = 6.626 \times 10^{-34}\,\text{J}\cdot\text{s})Photon Energy
  • KEmax=hfφKE_{max} = hf - \varphiEinstein Photoelectric Equation
  • eVstop=KEmaxeV_{stop} = KE_{max}Stopping Voltage relation
  • fthreshold=φhf_{threshold} = \frac{\varphi}{h}Threshold Frequency

Frequently asked questions

Why does increasing light intensity below the threshold frequency produce NO photoelectrons?
You can work it out this way: think about what a single photon's energy depends on.
Sodium has a work function φ = 2.3 eV (this sim's sodium preset). What is the minimum threshold frequency?
You can work it out this way: use f_threshold = φ/h, convert φ to joules first.
Light of f = 5 × 10¹⁴ Hz shines on a surface with work function φ = 4.5 eV (this sim's second preset). Are photoelectrons emitted?
You can work it out this way: calculate photon energy E = hf and compare with φ.