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Chapter 34: Problem 22

A microwave oven uses electromagnetic radiation at \(2.4 \mathrm{GHz}\). (a) What's the energy of each microwave photon? (b) At what rate does a \(900-\) W oven produce photons?

Short Answer

Expert verified
The energy of each microwave photon is approximately \(1.5904 \times 10^{-24} \, \mathrm{J}\). The \(900-\mathrm{W}\) oven produces photons at a rate of approximately \(5.657 \times 10^{26} \, \mathrm{s}^{-1}\) or \(5.657 \times 10^{26}\) photons per second.

Step by step solution

01

Calculate the photon's energy

In order to determine the energy of one photon, utilize the formula \( E = h \times f \), where \( h \) is Planck鈥檚 constant (\(6.626 \times 10^{-34} \mathrm{Js} \)) and \( f \) is the frequency of the radiation. Here, the frequency is provided in GHz, thus it needs to be converted to Hz by multiplying by \(10^9\). Thus, the formula becomes \( E = 6.626 \times 10^{-34} \times 2.4 \times 10^9 \) which gives \( E = 1.5904 \times 10^{-24} \, \mathrm{J} \).
02

Calculate the Photon Production Rate

The rate of photon production can be calculated by dividing the total power of the oven \( P \) by the energy of each photon \( E \). This yields the formula: \( \text{rate} = P / E \). Substitute the values from the problem: \( \text{rate} = 900 / 1.5904 \times 10^{-24} \), which simplifies to \( \text{rate} = 5.657 \times 10^{26} \, \mathrm{s}^{-1} \), meaning the oven produces approximately \( 5.657 \times 10^{26} \) photons per second.

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