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The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is 1.65 eV) will be
Options
(a) 4×10¹⁰ Hz
(b) 4×10¹⁴ Hz
(c) 4×10¹¹ Hz
(d) 4×10¹⁵ Hz
Correct Answer:
4×10¹⁴ Hz
Explanation:
Maximum wavelength (in Å) = 12375 / ɸ (in eV)
⇒ λ₀ = 12375 / 1.65 = 7,500 Å = 7.5 × 10⁻⁷ m
Hence minimum frequency = ʋ₀ = c / λ₀
⇒ ʋ₀ = (3 × 10⁸) / (7.5 × 10⁻⁷) = 4 × 10¹⁴ Hz
Related Questions: - The dual nature of light is exhibited by
- A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 secs
- The heart of a man pumps 5 litres of blood through the arteries per minute at
- A wire in the form of a circular loop of one turn, carrying a current, produces
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Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The dual nature of light is exhibited by
- A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 secs
- The heart of a man pumps 5 litres of blood through the arteries per minute at
- A wire in the form of a circular loop of one turn, carrying a current, produces
- There is a ring of radius r having linear charge densityλ and rotating with a uniform
Topics: Dual Nature of Matter and Radiation (150)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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