The ratio of the power of a light source S₁ to that the light source S₂ is 2 . S₁ is emitting 2 × 10¹⁵ photons per second at 600 nm . If the wavelength of the source S₂ is 300 nm , then the number of photons per second emitted by S₂ is __________ × 10¹⁴.

Solution

Physics sir · 4 steps

Power is photons per second times energy per photon, and a 300 nm photon carries twice the energy.

1. Power from photons. Each photon carries energy hc/λ. A source emitting n photons per second therefore has power P = nhc/λ.

   P₁/P₂ = (n₁/λ₁)/(n₂/λ₂) = (n₁ λ₂)/(n₂ λ₁)

2. Put in numbers. 2 = ((2 × 10¹⁵) × 300)/(n₂ × 600) = 10¹⁵/n₂, so n₂ = 5 × 10¹⁴ photons per second.
3. Check it. S₂ has half the power, and each of its photons has twice the energy. So it emits a quarter as many photons: (2 × 10¹⁵)/4 = 5 × 10¹⁴.
4. So. The number of photons per second from S₂ is 5 × 10¹⁴, so the answer is 5.

The common slip: 20. Taking photon energy as proportional to λ instead of 1/λ gives 2 × 10¹⁵, that is 20, not 5.

The idea to keep. Photon rate is n = Pλ/hc, so n ∝ Pλ. Halve the power and halve the wavelength, and the photon count drops to a quarter.

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