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)

The correct option is 10¹⁶.

To find the average number of photons emitted per second by the monochromatic light source, we can follow these steps:

Step 1: Understand the relation between power, photon energy, and emission rate
Power ($P$) is defined as the total energy delivered per second. If $n$ is the number of photons emitted per second and $E$ is the energy of a single photon, then:
$P=n\times E$
Rearranging the formula to solve for $n$:
$n=\frac{P}{E}$

Step 2: Calculate the energy of a single photon ($E$)
The energy of a photon is given by Planck's equation:
$E=\frac{hc}{\lambda }$
where:
• Planck's constant, $h=6.6\times {10}^{-34}J\; s$
• Speed of light, $c=3\times {10}^{8}m/s$
• Wavelength of light, $\lambda =600nm=600\times {10}^{-9}m=6\times {10}^{-7}m$

Substituting these values into the equation:
$E=\frac{(6.6\times {10}^{-34})\times (3\times {10}^8)}{6\times {10}^{-7}}$
$E=\frac{1.98\times {10}^{-25}}{6\times {10}^{-7}}$
$E=3.3\times {10}^{-19}Joules$

Step 3: Calculate the number of photons emitted per second ($n$)
Given the power delivered is $P=3.3\times {10}^{-3}W$:
$n=\frac{3.3\times {10}^{-3}}{3.3\times {10}^{-19}}$
$n={10}^{-3-(-19)}$
$n={10}^{16}$

Therefore, the average number of photons emitted per second is 10¹⁶.