Photoelectric Effect – What controls electron emission? A clean metal surface is exposed to light and emits electrons (photoelectric emission). Which factor determines whether electrons are emitted and thus initiates the effect?

Introduction / Context:
The photoelectric effect established the quantum nature of light. When photons strike a metal, electrons may be emitted if the photon energy exceeds the material work function. This phenomenon underpins photodetectors, solar cells (at junctions), and many optical sensors.

Given Data / Assumptions:

• Metal has a work function phi (energy required to liberate an electron).
• Photon energy is E = h * f, where h is Planck’s constant and f is frequency.
• Clean surface, negligible oxide layers for the conceptual explanation.

Concept / Approach:

Emission occurs only if photon energy exceeds the threshold: h * f ≥ phi. Thus, the decisive factor for emission onset is frequency (or equivalently wavelength). Intensity determines the number of incident photons per second and therefore the emission rate (photocurrent) once the threshold is surpassed, but cannot trigger emission if the frequency is too low.

Step-by-Step Solution:

Verification / Alternative check:

Classic experiments show instantaneous emission above threshold frequency and zero emission below it even under very high intensity at low frequency.

Why Other Options Are Wrong:

Intensity alone cannot cause emission below threshold. “Both equally” misstates the trigger condition. “Neither” ignores the established frequency threshold. Work function matters but the variable under experimental control that decides emission is light frequency crossing that threshold.

Common Pitfalls:

Confusing current magnitude (intensity-dependent) with emission occurrence (frequency-dependent); neglecting wavelength–frequency relation.

Final Answer:

Frequency of the light