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Question

Physics Question on communication systems

A 10 kW transmitter emits radio waves of wavelength 500 m. The number of photons emitted per second by the transmitter is of the order of

A

103710^{37}

B

103110^{31}

C

102510^{25}

D

104310^{43}

Answer

103110^{31}

Explanation

Solution

Power =nhcλ= \frac{nhc}{\lambda} (where, n = no. of photons per second) ⇒n=10×103×5006.6×10−34×3×108≃1031\Rightarrow\quad n = \frac{10\times10^{3}\times500}{6.6\times10^{-34}\times3\times10^{8}} \simeq 10^{31}