What is the threshold frequency f_0 for a metallic surface with a work function of Phi = 3.2 eV? (h = 4.14 10^(-15) eVs)
Answer and explanation
Correct answer: A. 7.73 * 10^14 Hz
- Step 1: Recall the photoelectric relation for threshold frequency: Phi = h * f_0.
- Step 2: Solve for f_0: f_0 = Phi / h.
- Step 3: Substitute values: f_0 = 3.2 eV / (4.14 10^(-15) eVs) approx 7.729 * 10^14 Hz.
Answer
7.73 * 10^14 Hz.
Why not the others: '4.82 10^14 Hz' uses an incorrect Planck constant; '1.32 10^15 Hz' inverts the quotient; '9.11 * 10^14 Hz' confuses with electron rest mass mantissa.
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