The work function of a certain photosensitive metallic surface is 2.40 eV. What is the threshold frequency of light capable of liberating photoelectrons from this metal? (Planck constant h = 6.63 × 10⁻³⁴ J·s, 1 eV = 1.6 × 10⁻¹⁹ J)
Answer and explanation
Correct answer: C. 5.79 × 10¹⁴ Hz
- Step 1: Convert work function to Joules: W0 = 2.40 × (1.6 × 10⁻¹⁹) = 3.84 × 10⁻¹⁹ J. Step 2. Photoelectric threshold relation: W0 = h f0 => f0 = W0 / h. Step 3. Calculate: f0 = (3.84 × 10⁻¹⁹) / (6.63 × 10⁻³⁴) ≈ 5.79 × 10¹⁴ Hz. Answer: 5.79 × 10¹⁴ Hz. Why not the others: 3.62 × 10¹⁴ Hz fails to convert eV to J properly; 7.24 × 10¹⁴ Hz and 4.15 × 10¹⁴ Hz represent calculation errors.
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