Physics · Dual Nature Of Radiation And Matter · NEET
n = P / E, where P is the power of the source in watts and E is the energy of one photon. Since E = hf = hc/lambda, you can also write n = P/(hf) or n = P*lambda/(hc). Here h = 6.63x10^-34 J s, c = 3x10^8 m/s. Power tells you the total energy given out each second; dividing by one photon's energy tells you how many photons that is.
Power P means the source gives out P joules of energy every second. Each photon carries a fixed energy E = hf. If the source gives out P joules and each packet is E joules, then the count of packets is P/E. That is exactly the number of photons per second. It is like sharing total money each second into equal coins.
Either works, they give the same answer. If frequency f is given, use E = hf. If wavelength lambda is given, use E = hc/lambda. So n = P/(hf) or n = P*lambda/(hc). Always convert wavelength to metres (1 nm = 10^-9 m) and keep power in watts (1 mW = 10^-3 W) before dividing.
For the same colour (same frequency), yes. If P doubles and E per photon is unchanged, then n = P/E doubles. But be careful: a red source and a blue source of the SAME power emit different photon rates, because a blue photon carries more energy, so fewer blue photons are needed to make the same power.
n comes out as photons per second (a pure count per second). Keep power in watts (J/s) and photon energy in joules, so the joules cancel and you are left with 1/second. A typical small source gives a very large n, like 10^15 to 10^20 photons per second, which is normal.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Energy of one photon E = hf = (6.63x10^-34)(6.0x10^14) = 3.98x10^-19 J. Number per second n = P/E = (2.0x10^-3)/(3.98x10^-19) = 5.0x10^15 photons per second. This is NCERT Example 11.1.
Yes. When wavelength is given instead of frequency, substitute E = hc/lambda into n = P/E to get n = P*lambda/(hc). Keep lambda in metres, P in watts, h = 6.63x10^-34 J s, c = 3x10^8 m/s.
No, but they are related. Intensity is power per unit area. The number of photons crossing a unit area per unit time depends on both intensity and photon energy. For the whole source, use n = P/E to get total photons per second emitted.
Each photon carries an extremely tiny energy (around 10^-19 J). So even a small power like a few milliwatts needs an enormous number of photons per second to add up to that power. Large answers like 10^15 are expected and correct.
