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The number of photons of wavelength of 540 nm emitted per second by an electric bulb of power 100 W is (given h=6×10⁻³⁴ J-s)
Options
(a) 100
(b) 1000
(c) 3×10²⁰
(d) 3×10¹⁸
Correct Answer:
3×10²⁰
Explanation:
Power = nhc / ( × t)
⇒ 100 = [n × (6 × 10⁻³⁴) × (3 × 10⁸)] / (540 × 10⁹ × 1)
⇒ n × 18 × 10⁻²⁶ / 540 × 10⁻⁹ = 100
⇒ n = 100 × 540 × 10⁻⁹ / 18 × 10⁻²⁶ = 3 x 10²⁰
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Topics: Dual Nature of Matter and Radiation
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Huygen’s wave theory allows us to know
- A heavy nucleus at rest breaks into two fragments which fly off with velocities
- Pure Si at 500 K has equal number of electron (nₑ) and hole (nₕ) concentrations
- The escape velocity of a body on an imaginary planet which is thrice the radius
- When the length of the vibrating segment of a sonometer wire is increased by
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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