Question

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Answer 1

Explanation:

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Answer 2

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Energy emitted by the transmitter per sec,Et=10kW

Energy of the wave, Ew=hc/λ 

where λ=300m

Plugging the values gives Ew=6.6×10−28

No of photons emitted per sec, n=Et/Ew=10000/6.6×10−28=1.5×1031