| ⇦ | 
| ⇨ |
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
Related Questions: - The velocity of radio-waves is 3×10⁵ km/s. The frequency corresponding to wavelength
- The energy of the em waves is of the order of 15 keV. To which part of the spectrum
- Amout of heat required to raise the temperature of a body through 1 K is called its
- If the magnetising field on a ferromagnetic material is increased, its permeability
- A particle is executing a simple harmonic motion. Its maximum acceleration
Topics: Dual Nature of Matter and Radiation
(150)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The velocity of radio-waves is 3×10⁵ km/s. The frequency corresponding to wavelength
- The energy of the em waves is of the order of 15 keV. To which part of the spectrum
- Amout of heat required to raise the temperature of a body through 1 K is called its
- If the magnetising field on a ferromagnetic material is increased, its permeability
- A particle is executing a simple harmonic motion. Its maximum acceleration
Topics: Dual Nature of Matter and Radiation (150)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply