A 5 watt LED bulb converts 80% of the power into light photons wavelength 660 nm. What is the number of photons emitted from the bulb in one second.
Correct answer: D. 1.3 x 10^19
- A. 5.8 x 10^34
- B. 7.5 x 10^18
- C. 6.6 x 10^7
- D. 1.3 x 10^19
Explanation
Photon energy formula: E=hc/λ , where h is planck's constantλ =660nmc= 3*108 m/sh= 6.626 × 10-34 JsE=6.626*10-34 * 3*108 / 660*10-9 (Js*m/s/m = J)E=3.01*10-19J(per photon)Total energy= 1sec*5watt*80/100(efficiency) = 4 ws = 4JNo. of photons = Total energy/E = 4J/3.01*10-19J = 1.3*1019
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Classical physics cannot explain blackbody radiation and related observations, leading to Planck’s quantum theory, in which energy is emitted or absorbed in discrete packets. Photons provide the particle model of light, with energy and momentum linked to frequency and wavelength, including the photoelectric effect and its threshold frequency.
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