⇦ | ![]() | ⇨ |
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:
- λ₁ and λ₂ are used to illuminate the slits. β₁ and β₂ are the corresponding fringe
- A ball moving with velocity 2 m/s collides head on with another stationary ball
- The half-life of radioactive element is 600 yr. The fraction of sample
- An electric motor operating on a 60 volt D.C supply draws a current of 10 amp.
- Electron in hydrogen atom first jumps from third excited state to second excited
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