用频率为 6×10¹⁴ Hz 的光照射某金属表面，恰好发生光电效应。已知普朗克常量 h = 6.63×10⁻³⁴ J·s，求该金属的逸出功（结果保留两位有效数字）。
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Correct answer
B. 4.0×10⁻¹⁹ J
Principle or equation
光电效应方程：hν = W0 + Ek，恰好发生光电效应时 Ek = 0，所以 W0 = hν。
Why this answer is correct
W0 = hν = 6.63×10⁻³⁴ × 6×10¹⁴ = 39.78×10⁻²⁰ = 3.978×10⁻¹⁹ J ≈ 4.0×10⁻¹⁹ J。
Example
若频率为 8×10¹⁴ Hz，则逸出功约为 5.3×10⁻¹⁹ J。
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