Explanation:
Photoelectric equation: work function φ = hν₀, where ν₀ is threshold frequency. Thus ν₀ = φ/h = 2.3 eV / (4.14×10⁻¹⁵ eV·s) ≈ 5.555×10¹⁴ Hz ≈ 5.6×10¹⁴ Hz. This direct application tests photoelectric effect fundamentals. Memory aid: 'ν₀ = φ/h; higher work function ⇒ higher threshold frequency'. Competitive exams frequently provide h in eV·s for such calculations. Always ensure units match: eV for φ and h to avoid conversion errors. This problem assesses numerical proficiency in modern physics applications.
Explanation:
Loudness correlates with intensity, which is proportional to square of amplitude. Frequency determines pitch. Larger amplitude → louder sound. Measured in decibels.
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