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 :

To solve this problem, we need to use the formula for the energy of a photon, which is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the light.

1. First, we need to convert the wavelengths from nanometers to meters. So, 300 nm = 300 x 10^-9 m and 500 nm = 500 x 10^-9 m.

2. Next, we calculate the energy of a photon for each wavelength using the formula E = hc/λ.

   For λ = 300 nm, E1 = (6.626 x 10^-34 J.s)(3 x 10^8 m/s) / (300 x 10^-9 m) = 6.626 x 10^-19 J

   For λ = 500 nm, E2 = (6.626 x 10^-34 J.s)(3 x 10^8 m/s) / (500 x 10^-9 m) = 3.976 x 10^-19 J

3. The number of photons emitted per second by a source is given by the power of the source divided by the energy of a photon. So,

   For the source with λ = 300 nm, N1 = 200 W / 6.626 x 10^-19 J = 3.02 x 10^21 photons/s

   For the source with λ = 500 nm, N2 = 200 W / 3.976 x 10^-19 J = 5.03 x 10^21 photons/s

4. The ratio of the number of photons emitted by the two sources is therefore N1/N2 = (3.02 x 10^21 photons/s) / (5.03 x 10^21 photons/s) = 0.60.

So, the ratio of the number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively is 0.60.