Physics: AM Radio Station Emits Photons/sec at 1000-kHz

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SUMMARY

An 84-kW AM radio station broadcasting at 1000 kHz emits photons at a rate determined by the energy of each photon and the total energy output. Using Planck's constant (h = 6.626E-34 Js), the energy of a single photon is calculated as Eν = hν = 6.626 × 10^-34 × 1,000,000 Hz = 6.626 × 10^-28 J. The total energy emitted per second is 8.4 × 10^4 Joules, leading to the calculation of the number of photons emitted per second as N = Es / Eν.

PREREQUISITES
  • Understanding of Planck's constant (h = 6.626E-34 Js)
  • Knowledge of photon energy calculations
  • Familiarity with frequency and its relation to energy
  • Basic principles of radio wave transmission
NEXT STEPS
  • Explore the concept of photon energy in different frequency ranges
  • Learn about the implications of photon emission in radio frequency applications
  • Investigate the relationship between power output and photon emission rates
  • Study the effects of frequency modulation on photon emission
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Physicists, electrical engineers, and students studying radio frequency transmission and photon behavior in electromagnetic waves.

yakin
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Physics: An 84-kW AM radio station broadcasts at 1000-kHz. How many photons are emitted each second by the transmitting antenna? (h=6.626E-32 Js)
 
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yakin said:
Physics: An 84-kW AM radio station broadcasts at 1000-kHz. How many photons are emitted each second by the transmitting antenna? (h=6.626E-32 Js)

Energy of a single photon with frequency $$\nu$$ is:

$$E_{\nu}=h \nu$$

where $$h$$ Plank's constant.

So for our 1000 kHz signal the energy of a single photon is:

$$E_{\nu}=h \nu= 6.626\times 10^{-32} \times 1000 \times 1000= 6.626\times 10^{-28}$$ J

The radio station is emitting $$E_s=8.4\times 10^4$$ Joules per second, so the number of photons emited per second is:

$$N=\frac{E_s}{E_{\nu}}$$

.
 

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