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The ratio of the power of a light source S₁ to that of 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.

Asked in JEE Main 24th Jan 2nd Shift 2025 · Counting photons

Answer: 5

Step-by-step solution

Given: (P₁)/(P₂)=2, n₁=2×10¹⁵ at 600 nm, λ₂=300 nm.

P=(nhc)/λ, so (P₁)/(P₂)=(n₁λ₂)/(n₂λ₁)=(n₁×300)/(n₂×600).

2=(n₁)/(2n₂), so n₂=(n₁)/4.

n₂=5×10¹⁴ per second, so the answer is 5.

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