/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 An AM radio station broadcasts a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An AM radio station broadcasts an electromagnetic wave with a frequency of \(665 \mathrm{kHz}\), whereas an \(\mathrm{FM}\) station broadcasts an electromagnetic wave with a frequency of 91.9 MHz. How many AM photons are needed to have a total energy equal to that of one FM photon?

Short Answer

Expert verified
138 AM photons are needed to have the same energy as one FM photon.

Step by step solution

01

Calculate the energy of one FM photon

The energy of a photon can be calculated using the formula \( E = h \cdot f \), where \( E \) is energy, \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{Js}\), and \( f \) is frequency. For the FM station with frequency \(91.9 \text{ MHz} = 91.9 \times 10^6 \text{ Hz}\), the energy of one FM photon is calculated as follows: \[ E_{FM} = (6.626 \times 10^{-34}) \cdot (91.9 \times 10^6) \approx 6.09 \times 10^{-26} \text{ J}. \]
02

Calculate the energy of one AM photon

Similarly, for the AM station with frequency \(665 \text{ kHz} = 665 \times 10^3 \text{ Hz}\), use the formula \( E = h \cdot f \) to find the energy of one AM photon: \[ E_{AM} = (6.626 \times 10^{-34}) \cdot (665 \times 10^3) \approx 4.41 \times 10^{-28} \text{ J}. \]
03

Calculate the number of AM photons

To find out how many AM photons are needed to match the energy of one FM photon, divide the energy of one FM photon by the energy of one AM photon: \[ \text{Number of AM photons} = \frac{E_{FM}}{E_{AM}} = \frac{6.09 \times 10^{-26}}{4.41 \times 10^{-28}} \approx 138. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Planck's constant
Planck's constant is a fundamental constant in physics, denoted by the symbol \( h \). It was introduced by Max Planck in the early 20th century and plays a crucial role in quantum mechanics. The value of Planck's constant is approximately \( 6.626 \times 10^{-34} \) Joule-seconds (Js). This small number indicates how minuscule quantum effects are compared to our everyday scale.

Planck's constant relates the energy of a photon to the frequency of its electromagnetic wave. The formula for photon energy is \( E = h \cdot f \), where \( E \) is energy, \( h \) is Planck's constant, and \( f \) is the frequency of the electromagnetic wave. This equation highlights one of the main ideas in quantum mechanics: energy is quantized, or divided into discrete packets, called quanta. In the context of this exercise, Planck's constant is used to calculate the energy of photons from both AM and FM radio stations.
Electromagnetic waves
Electromagnetic waves are waves of energy that transfer through space at the speed of light, which is approximately \( 3 \times 10^8 \) meters per second. These waves are composed of oscillating electric and magnetic fields that propagate at right angles to each other.

Electromagnetic waves cover a vast spectrum of frequencies and wavelengths, which are inversely related: as one increases, the other decreases. This spectrum includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays, each with its own frequency and wavelength range.
  • AM and FM radio waves are types of electromagnetic waves, distinguished by their frequency ranges.
  • AM radio uses amplitude modulation with frequencies typically in the kilohertz (kHz) range.
  • FM radio uses frequency modulation with frequencies in the megahertz (MHz) range.
These radio waves can travel considerable distances, making them ideal for broadcasting information over long distances like radio signals in our exercise.
Frequency calculation
Frequency calculation is a critical part of determining the energy of photons in any electromagnetic wave. Frequency, denoted by \( f \), represents how many waves pass a certain point per second, measured in hertz (Hz), which is equivalent to cycles per second.

In the exercise, the FM and AM stations broadcast at frequencies of 91.9 MHz and 665 kHz, respectively. To compute photon energy:
  • Convert frequencies to hertz: 91.9 MHz = \( 91.9 \times 10^6 \) Hz and 665 kHz = \( 665 \times 10^3 \) Hz.
  • Apply the formula \( E = h \cdot f \), inserting the corresponding frequency and Planck's constant \( 6.626 \times 10^{-34} \) Js.
In this exercise, calculating the frequency helps us to determine how many AM photons are needed to match the energy of an FM photon. By understanding how frequency affects photon energy, students can gain deeper insights into the nature of electromagnetic waves and their applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a Young's double-slit experiment that uses electrons, the angle that locates the first-order bright fringes is \(\theta_{\mathrm{A}}=1.6 \times 10^{-4}\) degrees when the magnitude of the electron momentum is \(p_{\mathrm{A}}=1.2 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) With the same double slit, what momentum magnitude \(p_{\mathrm{B}}\) is necessary so that an angle of \(\theta_{\mathrm{B}}=4.0 \times 10^{-4}\) degrees locates the first-order bright fringes?

Radiation of a certain wavelength causes electrons with a maximum kinetic energy of \(0.68 \mathrm{eV}\) to be ejected from a metal whose work function is 2.75 eV. What will be the maximum kinetic energy (in eV) with which this same radiation ejects electrons from another metal whose work function is 2.17 eV?

The work function of a metal surface is \(4.80 \times 10^{-19}\) J. The maximum speed of the electrons emitted from the surface is \(v_{\mathrm{A}}=7.30 \times 10^{5} \mathrm{m} / \mathrm{s}\) when the wavelength of the light is \(\lambda_{\mathrm{A}} .\) However, a maximum speed of \(v_{\mathrm{B}}=5.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) is observed when the wavelength is \(\lambda_{\mathrm{B}} .\) Find the wavelengths \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}.\)

ssm A particle has a de Broglie wavelength of \(2.7 \times 10^{-10} \mathrm{m}\). Then its kinetic energy doubles. What is the particle's new de Broglie wavelength, assuming that relativistic effects can be ignored?

Some scientists have suggested that spacecraft with sails of the kind described in Conceptual Example 4 could be propelled by lasers. Suppose that such a sail is constructed of a highly reflective material thin enough so that one square meter of the sail has a mass of just \(3.0 \times 10^{-3} \mathrm{kg} .\) The sail is to be propelled by an ultraviolet laser beam (wavelength \(=225 \mathrm{nm}\) ) that will strike its surface perpendicularly. (a) Use the impulse-momentum theorem (Section 7.1\()\) to determine the number of photons per second that must strike each square meter of the sail in order to cause an acceleration of \(9.8 \times 10^{-6} \mathrm{m} / \mathrm{s}^{2},\) which is one million times smaller than the gravitational acceleration at the earth's surface. Assume that no other forces act on the sail, and that all the incident photons are reflected. (b) Determine the intensity (power per unit area) that the laser beam must have when it strikes the sail.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.