Monochromatic light of frequency $6 \times 10^{14} \mathrm{~Hz}$ is produced by a laser. The power emitted is $2 \times 10^{-3} \mathrm{~W}$.

How many photons per second on an average, are emitted by the source ?

(Given $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$ )

<p>To find out the number of photons emitted per second by the laser, we can use the relationship between the energy of a single photon, the total energy emitted per second (power), and the number of photons emitted per second. The energy $E$ of a single photon is given by Planck's equation:</p> <p>$E = hf$</p> <p>where:</p> <ul> <li>$ h $ is Planck's constant ($ 6.63 \times 10^{-34} \mathrm{Js} $), and</li> <li>$ f $ is the frequency of the light ($ 6 \times 10^{14} \mathrm{Hz} $).</li> </ul> <p>Let's first calculate the energy of one photon:</p> <p>$E = (6.63 \times 10^{-34} \mathrm{Js}) \times (6 \times 10^{14} \mathrm{Hz})$</p> <p>$E = 3.978 \times 10^{-19} \mathrm{J}$</p> <p>The power ($ P $) emitted by the laser is the total energy emitted per second,</p> <p>$$ P = E_{\text{total per second}} = 2 \times 10^{-3} \mathrm{W} = 2 \times 10^{-3} \mathrm{J/s} $$</p> <p>The number of photons ($ N $) emitted per second can be found by dividing the total energy emitted per second by the energy of one photon:</p> <p>$N = \frac{P}{E}$</p> <p>Substitute the values we have:</p> <p>$N = \frac{2 \times 10^{-3} \mathrm{J/s}}{3.978 \times 10^{-19} \mathrm{J}}$</p> <p>$N = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}}$</p> <p>$N = 5.03 \times 10^{15} \text{ photons per second}$</p> <p>The number of photons emitted per second is approximately $5 \times 10^{15}$. Therefore, the correct answer, rounded to one significant figure, is:</p> <p>Option A: $5 \times 10^{15}$</p>