A metal has a work function of 2.3 eV. What is its threshold frequency (h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C)?
Correct answer D. 5.6 × 10¹⁴ Hz
Explanation
f₀ = φ/h = (2.3 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 3.68 × 10⁻¹⁹/6.63 × 10⁻³⁴ ≈ 5.6 × 10¹⁴ Hz.
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