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A 200 W sodium street lamp emits yellow light of wavelength 0.6 μm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
- 1.5 × 10^20
- 6 × 10^18
- 62 × 10^20
- 3 × 10^19
Correct answer: 3 × 10^19
Solution
The power converted to light is 25% of 200 W, i.e., 50 W. The energy of one photon is given by E = hc/λ. Substituting h = 6.63 × 10^-34 J·s, c = 3 × 10^8 m/s, and λ = 0.6 × 10^-6 m, we get E ≈ 3.315 × 10^-19 J. The number of photons emitted per second is Power/Energy per photon = 50/3.315 × 10^-19 ≈ 3 × 10^19.
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