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 .

Step-by-step solution

Since power emitting by a source is given as aligned & = Total energy emitted time & = (E₁ photon ) Number of photons (N) t & P₁= (E₁ ) n aligned aligned & P ₁ P ₂ = ( E ₁ ) n ₁ ( E ₂ ) n ₂ = ( hC ₁ ) n ₁ ( hC ₂ ) n ₂ & P ₁ P ₂ = ( ₂ ₁ ) n ₁ n ₂ aligned Substituting the given values aligned & 2= ( 300 600 ) 2 10¹⁵ n ₂ & n ₂= 1 2 10¹⁵=5 10¹⁴ Photon / sec aligned