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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 \(1.59 \times 10^{-24} \mathrm{J}\) and the oven generates about \(5.66 \times 10^{28}\) photons per second.

Step by step solution

01

Determine the Energy of each Photon

Firstly, as per Planck's formula, energy (E) = h * v where h is Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)) and v is frequency (2.4 GHz = \(2.4 \times 10^{9} \mathrm{Hz}\)). Hence, E = \(6.626 \times 10^{-34} \mathrm{Js}\) * \(2.4 \times 10^{9} \mathrm{Hz}\).
02

Calculate the Energy

Multiply the values to obtain the energy of each photon. E = \(6.626 \times 10^{-34} \mathrm{Js}\) * \(2.4 \times 10^{9} \mathrm{Hz}\) = \(1.59 \times 10^{-24} \mathrm{J}\).
03

Calculate the Rate of Photon Production

The rate at which photons are produced (R) = total power of oven/energy of each photon. Substituting the given values, R = 900 W / \(1.59 \times 10^{-24} \mathrm{J}\).
04

Calculate the Production Rate

Calculate the rate to determine the number of photons produced per second. R = 900/ \(1.59 \times 10^{-24}\) = \(5.66 \times 10^{28}\) photons per second.

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