AP Physics 2mediummcq1 pt
The work function of a certain metal is 4.0 eV. What is the minimum frequency of light that can cause photoelectric emission from this metal?
A.B) 6.0 × 10^14 Hz
B.C) 9.7 × 10^14 Hz
C.D) 1.2 × 10^15 Hz
D.A) 3.0 × 10^14 Hz
Explanation
Core Concept
The minimum frequency (threshold frequency) occurs when the photon energy equals the work function: hf = φ. Using h = 4.14 × 10^-15 eV·s, we get f = φ/h = 4.0 eV / (4.14 × 10^-15 eV·s) ≈ 9.7 × 10^14 Hz.
Correct Answer
BC) 9.7 × 10^14 Hz
Practice more AP Physics 2 questions with AI-powered explanations
Practice Unit 7: Quantum, Atomic, and Nuclear Physics Questions →