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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength Calculation
Calculating the wavelength of microwaves involves understanding the relationship between the speed of light, frequency, and wavelength. The speed of light, denoted as \(c\), is a fundamental constant with a value of \(3 \times 10^8 \text{ m/s}\). Microwaves in a typical oven operate at a frequency \(f\), which is given as \(2.2 \times 10^9 \text{ Hz}\).

Using the formula \(c = fλ\) (where \(λ\) is the wavelength), we can rearrange it to find \(λ\) by \(λ = \frac{c}{f}\). Substituting in the values:

  • Speed of light, \(c = 3 \times 10^8 \text{ m/s}\)
  • Frequency, \(f = 2.2 \times 10^9 \text{ Hz}\)

we get:

\[λ = \frac{3 \times 10^8}{2.2 \times 10^9} = 1.36 \times 10^{-1} \text{ meters}\]

This means the wavelength of the microwaves used in the oven is about 13.6 cm, which illustrates how microwaves fall within the millimeter to meter range.
Energy Calculation
The energy of a photon is related to its frequency through Planck's equation \(E = hf\). Here, \(E\) is the energy of the photon, \(h\) is Planck's constant \(6.63 \times 10^{-34} \text{ Js}\), and \(f\) is the frequency \(2.2 \times 10^9 \text{ Hz}\).

To find the energy of a microwave photon, use the given values:

  • Planck's constant, \(h = 6.63 \times 10^{-34} \text{ Js}\)
  • Frequency, \(f = 2.2 \times 10^9 \text{ Hz}\)

Substituting these into the formula \(E = hf\):

\[E = (6.63 \times 10^{-34})(2.2 \times 10^9) = 1.46 \times 10^{-24} \text{ Joules}\]

This shows that the energy of a single microwave photon is very small, about \(1.46 \times 10^{-24} \text{ J}\). This is why many photons are needed to heat food effectively, despite each having such a tiny amount of energy.
Physics Constants
Physics constants are fundamental values used in a multitude of calculations. In this exercise, two key constants are essential: the speed of light \(c\) and Planck's constant \(h\).

  • The speed of light \(c = 3 \times 10^8 \text{ m/s}\) is crucial in wave calculations and relates the wavelength and frequency of electromagnetic waves.
  • Planck's constant \(h = 6.63 \times 10^{-34} \text{ Js}\) connects the energy of photons with their frequency, a fundamental principle of quantum mechanics.

Understanding these constants helps simplify solving problems involving waves and energy. For example, the speed of light connects the concepts of wavelength and frequency, allowing us to determine one if we have the other. Meanwhile, Planck's constant helps us find the energy of a photon once its frequency is known.

Both constants are used extensively in various areas of physics, reflecting their significance in explaining natural phenomena and enabling technological advancements, such as the development of microwave ovens to efficiently heat food using electromagnetic waves.

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Most popular questions from this chapter

Solar radiation is incident on the outer surface of a spaceship at a rate of \(400 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\). The surface has an absorptivity of \(\alpha_{s}=0.10\) for solar radiation and an emissivity of \(\varepsilon=0.6\) at room temperature. The outer surface radiates heat into space at \(0 \mathrm{R}\). If there is no net heat transfer into the spaceship, determine the equilibrium temperature of the surface

A long metal bar \(\left(c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=\right.\) \(7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) is being conveyed through a water bath to be quenched. The metal bar has a cross section of \(30 \mathrm{~mm} \times 15 \mathrm{~mm}\), and it enters the water bath at \(700^{\circ} \mathrm{C}\). During the quenching process, \(500 \mathrm{~kW}\) of heat is released from the bar in the water bath. In order to prevent thermal burn on people handling the metal bar, it must exit the water bath at a temperature below \(45^{\circ} \mathrm{C}\). A radiometer is placed normal to and at a distance of \(1 \mathrm{~m}\) from the bar to monitor the exit temperature. The radiometer receives radiation from a target area of \(1 \mathrm{~cm}^{2}\) of the bar surface. Irradiation signal detected by the radiometer is used to control the speed of the bar being conveyed through the water bath so that the exit temperature is safe for handing. If the radiometer detects an irradiation of \(0.015 \mathrm{~W} / \mathrm{m}^{2}\), determine the speed of the bar bar can be approximated as a blackbody.

A small circular plate with a surface \(A_{1}\) can be approximated as a blackbody. A radiometer with a surface \(A_{2}\) is placed normal to the direction of viewing from the plate at a distance \(L\). Would the irradiation on the radiometer from the plate decrease if the distance \(L\) is doubled, and if so, by how much?

Explain why the sky is blue and the sunset is yellow-orange.

Consider a 4-cm-diameter and 6-cm-long cylindrical rod at \(1000 \mathrm{~K}\). If the emissivity of the rod surface is \(0.75\), the total amount of radiation emitted by all surfaces of the rod in \(20 \mathrm{~min}\) is (a) \(43 \mathrm{~kJ}\) (b) \(385 \mathrm{~kJ}\) (c) \(434 \mathrm{~kJ}\) (d) \(513 \mathrm{~kJ}\) (e) \(684 \mathrm{~kJ}\)

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