Photon Emission Rate Of A Laser
A small laser diode of output power 2 mW emits a steady beam of monochromatic red light of wavelength 660 nm. Approximately how many photons does the diode emit each second?
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Solution
6.6×1015
The number of photons emitted per second is the total radiated power divided by the energy of one photon, since each photon carries an indivisible quantum of energy. The single-photon energy is E=λhc=660×10−96.6×10−34×3×108=3.0×10−19 J, equivalent to about 1.88 eV. The emission rate is then N=EP=3.0×10−192×10−3≈6.6×1015 photons per second. The value 3.3×1015 wrongly doubles the photon energy. The value 1.3×1016 halves it. The value 2.0×1018 ignores the photon energy and divides power by the electronic charge instead. This calculation makes vivid how enormously many photons constitute even a feeble milliwatt beam, which is why the granular, quantised nature of light is hidden in everyday optics, a point underscored in the NCERT dual-nature chapter. A plausibility check confirms the answer is a very large but finite number, consistent with light appearing perfectly continuous to our senses despite being made of discrete quanta. It is instructive to appreciate that this staggering rate of roughly seven thousand million million photons each second is what makes the beam appear perfectly smooth and continuous to the eye and to ordinary detectors. Only with extremely sensitive single-photon counters operated at very low light levels does the discrete, grainy arrival of individual quanta become directly observable, confirming the particle character of light.
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About This Question
- Subject
- physics
- Chapter
- dual nature of radiation and matter
- Topic
- photon emission rate of a laser
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
6.6×1015
The number of photons emitted per second is the total radiated power divided by the energy of one photon, since each photon carries an indivisible quantum of energy. The single-photon energy is E=λhc=660×10−96.6×10−34×3×108=3.0×10−19 J, equivalent to about 1.88 eV. The emission rate is then N=EP=3.0×10−192×10−3≈6.6×1015 photons per second. The value 3.3×1015 wrongly doubles the photon energy. The value 1.3×1016 halves it. The value 2.0×1018 ignores the photon energy and divides power by the electronic charge instead. This calculation makes vivid how enormously many photons constitute even a feeble milliwatt beam, which is why the granular, quantised nature of light is hidden in everyday optics, a point underscored in the NCERT dual-nature chapter. A plausibility check confirms the answer is a very large but finite number, consistent with light appearing perfectly continuous to our senses despite being made of discrete quanta. It is instructive to appreciate that this staggering rate of roughly seven thousand million million photons each second is what makes the beam appear perfectly smooth and continuous to the eye and to ordinary detectors. Only with extremely sensitive single-photon counters operated at very low light levels does the discrete, grainy arrival of individual quanta become directly observable, confirming the particle character of light.
This medium difficulty physics question is from the chapter dual nature of radiation and matter, covering the topic of photon emission rate of a laser. It appeared in the 2025 exam.
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