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 answer is 10¹⁶.

To find the number of photons emitted per second, we first need to calculate the energy of a single photon. The energy ($E$) of a photon is given by the formula:

$E=\frac{hc}{\lambda }$

where:

• $h$ is Planck's constant ($6.6\times {10}^{-34}\text{J·s}$)
• $c$ is the speed of light in a vacuum ($3\times {10}^8\text{m/s}$)
• $\lambda$ is the wavelength of the light ($600\text{nm}=600\times {10}^{-9}\text{m}$)

Substitute these values into the energy formula:

$E=\frac{(6.6\times {10}^{-34})\times (3\times {10}^8)}{600\times {10}^{-9}}$

$E=\frac{19.8\times {10}^{-26}}{6\times {10}^{-7}}$

$E=3.3\times {10}^{-19}\text{Joules}$

Next, we use the given power of the source. Power ($P$) is defined as the total energy delivered per second. Therefore, the number of photons emitted per second ($n$) is the total power divided by the energy of a single photon:

$n=\frac{P}{E}$

Given the power $P=3.3\times {10}^{-3}\text{W}$ (which is Joules per second), we find:

$n=\frac{3.3\times {10}^{-3}}{3.3\times {10}^{-19}}$

$n={10}^{-3-(-19)}$

$n={10}^{16}$

Thus, the source emits ${10}^{16}$ photons per second.