Answer :
Final answer:
The question asks for the energy of photons released by an AM radio station. We used Planck's equation and converted the station's broadcast frequency from kHz to Hz to find the energy, which is roughly 6.70 × 10-24 joules per photon.
Explanation:
The energy of the photons emitted by the AM radio station can be calculated using Planck's equation which is E = hv, where E is the energy, h is Planck's constant (6.63 × 10-34 J.s), and v is the frequency. But since in the problem, frequency is given in kilohertz (kHz), we need to convert it to hertz by multiplying it by 103. So, the frequency of the AM radio station's broadcast is 1010 x 103Hz.
Now substitute h and v into Planck's equation:
E = 6.63 × 10-34 J.s × 1010 x 103s-1
Therefore, the energy of the photons emitted by the AM radio station is approximately 6.70 × 10-24 joules per photon.
To calculate the energy of the photons emitted by the AM radio station, we can use Planck's equation: E = hv, where E is the energy, h is Planck's constant (6.63 × 10^-34 J·s), and v is the frequency of the radio signal. The frequency of the AM radio station is 1010 kHz, which can be converted to 1010 × 10^3 Hz. Plugging in the values, we get: E = (6.63 × 10^-34 J·s) × (1010 × 10^3 Hz) = 6.70 × 10^-28 J.
Learn more about Planck's equation here:
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