The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3×10⁻³ watt will be : (h=6.6×10⁻³⁴ Js)

To find the number of photons emitted per second by the source of monochromatic light, we can use the relation between power, energy of a single photon, and the rate of emission of photons.

The total power $P$ delivered by the source is the total energy emitted per second. If $n$ is the number of photons emitted per second and $E$ is the energy of a single photon, then:
$P=n\times E$
Therefore, the number of photons emitted per second is:
$n=\frac{P}{E}$

According to Planck's quantum theory, the energy $E$ of a single photon of wavelength $\lambda$ is given by:
$E=\frac{hc}{\lambda }$
where:
- $h$ is Planck's constant = $6.6\times {10}^{-34}\text{Js}$
- $c$ is the speed of light in vacuum ≈ $3\times {10}^8\text{m/s}$
- $\lambda$ is the wavelength of light = $600\text{nm}=600\times {10}^{-9}\text{m}=6\times {10}^{-7}\text{m}$

First, let's calculate the energy of a single photon:
$E=\frac{\left(6.6\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{6\times {10}^{-7}}$
Simplifying the values:
$E=\frac{19.8\times {10}^{-26}}{6\times {10}^{-7}}$
$E=3.3\times {10}^{-19}\text{Joules}$

Now, substitute the values of power $P=3.3\times {10}^{-3}\text{W}$ and photon energy $E$ back into the formula for $n$:
$n=\frac{3.3\times {10}^{-3}}{3.3\times {10}^{-19}}$
$n={10}^{-3-\left(-19\right)}$
$n={10}^{16}{\text{s}}^{-1}$

Thus, the average number of photons emitted per second by the source is 10¹⁶.