A 100 W point source emits monochromatic light of wavelength 6000 $A^o$
Calculate the photon flux (in SI unit) at a distance of 5m from the source. Given $h = 6.6 \times 10 ^{34}$ Js and $C = 3 \times 10^8 ms^{-1}$.

Hint: Find the intensity of the photon using the intensity formula of the photon and then find the value of the energy of the photon with the given wavelength and finally find the value of the flux of the photon using the formula as $\dfrac{I}{E}$ .

Complete step-by-step solution:
Given the power of source P = 100 W
The wavelength of the emitted monochromatic light $\lambda= 6000 A^o$
Speed of light in vacuum C = $3\times 10^8 ms^{-1}$
Planck's constant h = $6.6\times 10^{-34} J$
Now we have to calculate the photon flux at a distance of 5 m from the source.
Let us find the intensity I
$I = \dfrac{P}{\text{surface area}} =\dfrac{P}{4\pi r^2}$ where r = 5 m
$I = \dfrac{100}{4\pi 5^2} = \dfrac{1}{\pi} \dfrac{W}{m^2}$
We know that photon flux is the number of photons passing normally per unit area per unit time.
Therefore photon flux = $I/E$
Where E = energy of photon = $\dfrac{hc}{\lambda} =\dfrac{ 6.6 \times 10^{-34} \times 3 \times 10^8}{ 6000 \times 10^{-10}} = 3.3 \times 10^{-19}$
So now we have got the energy of the photon and now we need to find the photon flux as below :
Photon flux = $\dfrac{I}{E} = \dfrac{\dfrac{1}{\pi} }{3.3 \times 10 ^{-19}} = 10 ^{18}$ photons $m^{-2}s^{-1}$.

Note: The possible mistake that one can make in this kind of problem is that we may take the photon intensity as the photon flux, which is wrong. There is a difference here. Intensity is the energy flux of the photon and the photon flux is the number of photons passing normally per unit area. So we need to take care of it.