Photon Flux And Power
A monochromatic lamp emits light of wavelength 600 nm at a radiant power of 3.98 watts, and an engineer estimates how many photons leave the lamp each second.
Select the correct option:
Solution
Approximately 1.2 × 10¹⁹ photons per second
The number of photons emitted per second is the total power divided by the energy per photon, a relation NCERT builds from E=λhc. First find the energy per photon: E=600×10−9(6.63×10−34)(3×108)=6×10−71.989×10−25=3.315×10−19 J. Then the photon rate is n=EP=3.315×10−193.98≈1.2×1019 photons per second. The enormous number reflects how little energy each photon carries. The value 1.2×1016 photons per second is wrong because it misplaces the exponent by three, likely from a wavelength conversion slip. The value 3.3×1019 photons per second is wrong because it inverts the arithmetic, effectively using energy per photon as the numerator. The value 1.2×1022 photons per second is wrong because it overshoots by three orders of magnitude. As stated in NCERT Class 12, Chapter 11, macroscopic light beams contain astronomically many photons, which is why their granular nature is normally hidden. The sheer scale of this number explains why the discrete, granular nature of light is normally invisible to us: with quintillions of photons per second, the beam appears perfectly smooth and continuous. Only at extremely low light levels, where photons arrive countably, does the underlying particle nature become directly apparent. A magnitude check confirms that dividing a few watts by a sub-attojoule photon energy sensibly yields around 1019 photons per second.
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About This Question
- Subject
- physics
- Chapter
- dual nature of matter and radiation
- Topic
- photon flux and power
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Approximately 1.2 × 10¹⁹ photons per second
The number of photons emitted per second is the total power divided by the energy per photon, a relation NCERT builds from E=λhc. First find the energy per photon: E=600×10−9(6.63×10−34)(3×108)=6×10−71.989×10−25=3.315×10−19 J. Then the photon rate is n=EP=3.315×10−193.98≈1.2×1019 photons per second. The enormous number reflects how little energy each photon carries. The value 1.2×1016 photons per second is wrong because it misplaces the exponent by three, likely from a wavelength conversion slip. The value 3.3×1019 photons per second is wrong because it inverts the arithmetic, effectively using energy per photon as the numerator. The value 1.2×1022 photons per second is wrong because it overshoots by three orders of magnitude. As stated in NCERT Class 12, Chapter 11, macroscopic light beams contain astronomically many photons, which is why their granular nature is normally hidden. The sheer scale of this number explains why the discrete, granular nature of light is normally invisible to us: with quintillions of photons per second, the beam appears perfectly smooth and continuous. Only at extremely low light levels, where photons arrive countably, does the underlying particle nature become directly apparent. A magnitude check confirms that dividing a few watts by a sub-attojoule photon energy sensibly yields around 1019 photons per second.
This medium difficulty physics question is from the chapter dual nature of matter and radiation, covering the topic of photon flux and power. It appeared in the 2025 exam.
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