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Chapter 7: Problem 98

AM radio stations broadcast at frequencies between \(530 \mathrm{kHz}\) and \(1700 \mathrm{kHz}\). (1 \(\mathrm{kHz}=10^{3} \mathrm{~s}^{-1}\).) For a station broadcasting at \(1.69 \times 10^{3} \mathrm{kHz}\), what is the energy of this radio wave? Note that Planck's constant is \(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\), and the speed of light is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The energy of the radio wave is approximately \(1.12 \times 10^{-27} \text{ J}\).

Step by step solution

01

Convert Frequency to Hertz

The frequency given is in kilohertz (kHz), which needs to be converted to Hertz (Hz) for calculation purposes. Since 1 kHz equals 10^3 Hz, the frequency can be converted by multiplying:\[1.69 \times 10^3 \text{ kHz} = 1.69 \times 10^6 \text{ Hz}\]
02

Use Planck's Equation for Energy

Planck's equation relates the energy (E) of a photon to its frequency (f) using the formula:\[E = h \cdot f\]where \(h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}\), and \(f = 1.69 \times 10^6 \text{ Hz}\) from Step 1. Substitute these into the equation to calculate the energy.
03

Calculate the Energy

Substitute the values into Planck's equation:\[E = (6.63 \times 10^{-34}) \times (1.69 \times 10^6)\]Perform the multiplication to determine the energy:\[E = 1.12147 \times 10^{-27} \text{ J}\]

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Key Concepts

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

Frequency Conversion
Converting frequency from one unit to another is often necessary in scientific calculations. In our exercise, the radio station frequency is initially given in kilohertz (kHz).
Kilohertz is a unit of frequency equivalent to 1,000 cycles per second or 1,000 Hertz (Hz).
To convert from kilohertz to Hertz, you multiply by 1,000.
For example, if a frequency is 1.69 kHz, you convert it by multiplying:
  • 1.69 kHz 脳 1,000 = 1.69 脳 103 Hz = 1.69 脳 106 Hz
This conversion is crucial when calculating energy using Planck's equation, as the frequency must be in Hertz.
Planck's Equation
Planck's equation is fundamental in quantum mechanics. It relates the energy of a photon to its frequency.
The equation is given by: \[E = h \cdot f\]where:
  • \(E\) is the energy of the photon.
  • \(h\) is Planck's constant, \(6.63 \times 10^{-34} \text{ J} \cdot \text{s}\).
  • \(f\) is the frequency of the photon.
The equation tells us that energy is directly proportional to frequency.
This means that a higher frequency results in higher energy for the photon.
It's important when studying electromagnetic waves and their energies.
Photon Energy
Photon energy is a key concept in understanding electromagnetic radiation. Photons are particles of light and other forms of electromagnetic radiation.
The energy of a photon is calculated using Planck's equation, where the frequency is needed in Hertz.
This energy is what impacts various forms of matter and is used in applications like radiation therapy or photovoltaic cells.
In our exercise, after converting the radio wave's frequency to Hertz, we used the equation:
  • \(E = h \times f\)
  • \(E = (6.63 \times 10^{-34}) \times (1.69 \times 10^6)\)
This calculation results in an energy of \(1.12147 \times 10^{-27} \text{ J}\).
Understanding photon energy is crucial in fields ranging from medicine to telecommunications.
Hertz to Kilohertz Conversion
Converting Hertz to kilohertz is useful for simplifying large numbers. In many cases, frequencies are very high, and working in kilohertz can make them easier to handle.
10 kilohertz signifies 10,000 Hertz (or 104 Hz).
To convert from Hertz to kilohertz, divide by 1,000:
  • For example, 1.69 million Hz is:
    • 1.69 脳 106 Hz 梅 1,000 = 1.69 脳 103 kHz
In our exercise, we began with kilohertz and converted to Hertz, ensuring accurate energy calculations from Planck's equation.
Understanding these conversions is foundational in fields dealing with electromagnetic waves, ensuring proper calculations and interpretations.

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Most popular questions from this chapter

An electron microscope employs a beam of electrons to obtain an image of an object. What energy must be imparted to each electron of the beam to obtain a wavelength of \(10.0 \mathrm{pm} ?\) Obtain the energy in electron volts \((\mathrm{eV})(1 \mathrm{eV}=1.602 \times\) \(\left.10^{-19} \mathrm{~J}\right)\)

The energy required to dissociate the \(\mathrm{Cl}_{2}\) molecule to \(\mathrm{Cl}\) atoms is \(239 \mathrm{~kJ} / \mathrm{mol} \mathrm{Cl}_{2}\). If the dissociation of a \(\mathrm{Cl}_{2}\) molecule were accomplished by the absorption of a single photon whose energy was exactly the quantity required, what would be its wavelength (in meters)?

An atom in its ground state absorbs a photon (photon 1 ), then quickly emits another photon (photon 2). One of these photons corresponds to ultraviolet radiation, whereas the other one corresponds to red light. Explain what is happening. Which electromagnetic radiation, ultraviolet or red light, is associated with the emitted photon (photon 2)?

What is the energy of a photon corresponding to radio waves of frequency \(1.365 \times 10^{6} / \mathrm{s}\) ?

Light, Energy, and the Hydrogen Atom a. Which has the greater wavelength, blue light or red light? b. How do the frequencies of blue light and red light compare? C. How does the energy of blue light compare with that of red light? d. Does blue light have a greater speed than red light? e. How does the energy of three photons from a blue light source compare with the energy of one photon of blue light from the same source? How does the energy of two photons corresponding to a wavelength of \(451 \mathrm{~nm}\) (blue light) compare with the energy of three photons corresponding to a wavelength of \(704 \mathrm{~nm}\) (red light)? f. A hydrogen atom with an electron in its ground state interacts with a photon of light with a wavelength of \(1.22 \times\) \(10^{-6} \mathrm{~m} .\) Could the electron make a transition from the ground state to a higher energy level? If it does make a transition, indicate which one. If no transition can occur, explain. g. If you have one mole of hydrogen atoms with their electrons in the \(n=1\) level, what is the minimum number of photons you would need to interact with these atoms in order to have all of their electrons promoted to the \(n=3\) level? What wavelength of light would you need to perform this experiment?
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