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Chapter 12: Problem 14

A microwave oven is designed to operate at a frequency of \(2.2 \times 10^{9} \mathrm{~Hz}\). Determine the wavelength of these microwaves and the energy of each microwave.

Short Answer

Expert verified
Answer: The wavelength of the microwaves is \(1.36 \times 10^{-1} \mathrm{m}\), and the energy of each microwave is \(1.46 \times 10^{-24} \mathrm{J}\).

Step by step solution

01

Recall the constants involved

In order to proceed, we will need the values for the speed of light (c) and Planck's constant (h). The speed of light is \(c=3 \times 10^8 \mathrm{m/s}\), and Planck's constant is \(h=6.63 \times 10^{-34} \mathrm{Js}\).
02

Calculate the wavelength using the frequency

Use the wavelength-frequency relationship \(c = fλ\) to determine the wavelength. We have the values for the speed of light (c) and frequency (f), and we need to find the wavelength (λ). By rearranging the formula as \(λ=\frac{c}{f}\) and substituting the given values, we obtain: \[ λ = \frac{3 \times 10^8 \mathrm{m/s}}{2.2 \times 10^9 \mathrm{Hz}}= 1.36 \times 10^{-1} \mathrm{m} \] So, the wavelength of these microwaves is \(1.36 \times 10^{-1} \mathrm{m}\).
03

Calculate the energy of each microwave

Use the energy-frequency relationship \(E = hf\) to determine the energy of each microwave. We have the values for Planck's constant (h) and frequency (f), so we can directly calculate the energy (E) as follows: \[ E=(6.63 \times 10^{-34} \mathrm{Js})(2.2 \times 10^9 \mathrm{Hz}) = 1.46 \times 10^{-24} \mathrm{J} \] Therefore, the energy of each microwave is \(1.46 \times 10^{-24} \mathrm{J}\).

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Most popular questions from this chapter

The sun can be treated as a blackbody at an effective surface temperature of \(10,400 \mathrm{R}\). Determine the rate at which infrared radiation energy \((\lambda=0.76-100 \mu \mathrm{m})\) is emitted by the sun, in Btu/h. \(\mathrm{ft}^{2}\).

We can see the inside of a microwave oven during operation through its glass door, which indicates that visible radiation is escaping the oven. Do you think that the harmful microwave radiation might also be escaping?

A surface at \(300^{\circ} \mathrm{C}\) has an emissivity of \(0.7\) in the wavelength range of \(0-4.4 \mu \mathrm{m}\) and \(0.3\) over the rest of the wavelength range. At a temperature of \(300^{\circ} \mathrm{C}, 19\) percent of the blackbody emissive power is in wavelength range up to \(4.4 \mu \mathrm{m}\). The total emissivity of this surface is (a) \(0.300\) (b) \(0.376\) (c) \(0.624\) (d) \(0.70\) (e) \(0.50\)

The temperature of the filament of an incandescent lightbulb is \(2500 \mathrm{~K}\). Assuming the filament to be a blackbody, determine the fraction of the radiant energy emitted by the filament that falls in the visible range. Also, determine the wavelength at which the emission of radiation from the filament peaks.

Define the properties reflectivity and transmissivity and discus the different forms of reflection.
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