Introduction / Context:
The photoelectric effect links photon energy to the kinetic energy of emitted electrons from a metal surface, described by Einstein's equation. It is foundational for modern quantum theory and optoelectronics.
Given Data / Assumptions:
- Metallic photocathode in vacuum.
- Incident light characterized by frequency f (or wavelength λ).
- Light intensity affects photon flux, not photon energy.
Concept / Approach:
Electron kinetic energy: KEmax = h f − φ, where φ is work function. Thus higher frequency (shorter wavelength) photons impart higher maximum kinetic energy to emitted electrons. Intensity mainly affects the number of emitted electrons (emission rate), not their maximum energy.
Step-by-Step Solution:
Step 1: Write Einstein relation: KEmax = h f − φ.Step 2: Increasing frequency (decreasing λ) increases KEmax and hence maximum velocity vmax.Step 3: Intensity changes photon count and thus emission rate, not KEmax.Step 4: Therefore, the correct statement is that maximum velocity increases with decreasing wavelength.
Verification / Alternative check:
Classical experiments: Stopping potential rises with frequency, independent of intensity, confirming dependence of KE on frequency.
Why Other Options Are Wrong:
A/E: Velocity (energy) depends on frequency, not intensity.B: Emission rate is directly, not inversely, proportional to intensity (if above threshold).D: Emission releases electrons only; metal does not emit free holes into vacuum.
Common Pitfalls:
Confusing photon energy (frequency) with photon flux (intensity); mixing photoelectric emission with photoconductive effects in solids.
Final Answer:
Maximum velocity of electron increases with decreasing wavelength
Discussion & Comments