Monochromatic light of frequency 6 times 10^{ | vedclass

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Question

(A) The power $P$ emitted by a source is given by the product of the number of photons emitted per second $(n)$ and the energy of a single photon $(E

Vedclass · Accepted Answer

$1 	imes 10^{16} 	ext{ photons per second}$

Vedclass · Answer

(A) The power $P$ emitted by a source is given by the product of the number of photons emitted per second $(n)$ and the energy of a single photon $(E = h
u)$.Given: Power $P = 4 	imes 10^{-3} 	ext{ W}$, frequency $
u = 6 	imes 10^{14} 	ext{ Hz}$, and Planck's constant $h = 6.63 	imes 10^{-34} 	ext{ Js}$.The formula is $P = n \cdot h
u$.Rearranging for $n$: $n = \frac{P}{h
u}$.Substituting the values: $n = \frac{4 	imes 10^{-3}}{6.63 	imes 10^{-34} 	imes 6 	imes 10^{14}}$.$n = \frac{4 	imes 10^{-3}}{39.78 	imes 10^{-20}} = \frac{4}{39.78} 	imes 10^{17} \approx 0.10055 	imes 10^{17} = 1.0055 	imes 10^{16}$.Rounding to the nearest significant figure, we get $n \approx 1 	imes 10^{16} 	ext{ photons per second}$.

Monochromatic light of frequency $6 \times 10^{14} \text{ Hz}$ is produced by a laser. The power emitted is $4 \times 10^{-3} \text{ W}$. How many photons per second on an average are emitted by the source? $[h = 6.63 \times 10^{-34} \text{ Js}]$

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