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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