Two sources of light emit with a power of 200 W. The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be :

Solution

Physics sir · 3 steps

Same power, so count photons by energy per photon: longer wavelength means cheaper photons, more of them.

1. Photons per second. Power is energy emitted per second. Each photon carries E = hc/λ, so the number emitted per second is the power divided by the energy of one photon.

   N = P/(hc/λ) = Pλ/hc

2. Take the ratio. Both sources have P = 200 W, so P, h and c cancel. That leaves N₁/N₂ = λ₁/λ₂ = (300 nm)/(500 nm) = 3/5.
3. So. The ratio of photons emitted is 3 : 5, option (b). The 500 nm source emits more photons because each one carries less energy.

The idea to keep. At fixed power, the number of photons per second is proportional to wavelength, N ∝ λ. Each high-energy (short-wavelength) photon uses up more of the energy budget.

Answer from the NTA final answer key. Solution by PrepWiser.