The work function of a metal is 2.3 eV. The threshold frequency for photoelectric emission is approximately: (h = 4.14×10⁻¹⁵ eV·s) MCQ with Answer and Explanation
The work function of a metal is 2.3 eV. The threshold frequency for photoelectric emission is approximately: (h = 4.14×10⁻¹⁵ eV·s)
A. 1.2×10¹⁵ Hz
B. 3.5×10¹⁴ Hz
C. 8.0×10¹⁴ Hz
D. 5.6×10¹⁴ Hz
Answer: Option D
Solution (By JKSSB Mock Tests)
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:
Rules: leading zeros not significant; trailing zero after decimal is significant. Digits 4,0,5,0 are significant ⇒ 4 sig figs. Memory tip: 'Start at first non-zero; include trailing zeros after decimal'. Measurement precision concept frequently tested in competitive exams.
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