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A 4,667 kHz AM radio station broadcasts with a power of 84 kW. How many photons does the transmitting antenna emit each second.

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Answer:

Step-by-step explanation:

Frequency( ν ) 4667 x 10³ Hz = 4.667 x 10⁶ Hz

Energy of one photon = hν [ h is plank's constant ]

= 6.6 x 10⁻³⁴ x 4.667 x 10⁶ = 30.8 x 10⁻²⁸ J

Power = 84 x 10³ J/s

No of photons emitted = Power / energy of one photon

= 84 x 10³ / 30.8 x 10⁻²⁸ =2.727 x 10³¹ per second .

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