Monochromatic light of frequency 6×1014 Hz is produced by a laser. If the power emitted is 2×10−3 W , then the number of photons per second on an average emitted by the source is given as P×1015 . Find the value of P. (h=6.63×10−34Js)

E=hνE=6.63×10−34×6×1014=3.978×10−19 JP=ETotal per second =2×10−3 W∴ No of photons =P Energy of each photon =2×10−33.978×10−19=5.03×1015 photon/sec

Monochromatic light of frequency 6×1014 Hz is produced by a laser. If the power emitted is 2×10−3 W , then the number of photons per second on an average emitted by the source is given as P×1015 . Find the value of P. (h=6.63×10−34Js)

E=hνE=6.63×10−34×6×1014=3.978×10−19 JP=ETotal per second =2×10−3 W∴ No of photons =P Energy of each photon =2×10−33.978×10−19=5.03×1015 photon/sec

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Monochromatic light of frequency $6 \times 10^{14} H z$ is produced by a laser. If the power emitted is $2 \times 10^{- 3} W$ , then the number of photons per second on an average emitted by the source is given as $P \times 10^{15}$ . Find the value of $P .$ $(h = 6.63 \times 10^{- 34} J s)$

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Detailed Solution

$\begin{array}{l} E = h ν \\ E = 6.63 \times 10^{- 34} \times 6 \times 10^{14} = 3.978 \times 10^{- 19} J \\ P = E_{Total per second} = 2 \times 10^{- 3} W \\ ∴ No of photons = \frac{P}{Energy of each photon} \\ = \frac{2 \times 10^{- 3}}{3.978 \times 10^{- 19}} = 5.03 \times 10^{15} photon / \sec \end{array}$