Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the wavelength of the electromagnetic waves used for cell phone communications in the 850 -MHz band?

Short Answer

Expert verified
Answer: The approximate wavelength of the electromagnetic waves used for cell phone communications in the 850 MHz band is 35.29 centimeters.

Step by step solution

01

Convert frequency to Hz

The frequency given is 850 MHz. To convert it to Hz, we need to multiply it by 10^6 (since there are one million Hz in a MHz). 850 MHz = 850 * 10^6 Hz = 8.50 * 10^8 Hz
02

Apply the formula to calculate the wavelength

We have the frequency (f) in Hz now, and the speed of light (c) is 3.00 * 10^8 m/s. We can apply the formula λ = c / f to find the wavelength. λ = (3.00 * 10^8 m/s) / (8.50 * 10^8 Hz)
03

Calculate the result

Dividing the speed of light by the frequency, we get: λ = 3.00 * 10^8 m/s / 8.50 * 10^8 Hz = 0.3529 * 10^0 m
04

Express the wavelength in an appropriate unit

The calculated wavelength is in meters. However, it's more appropriate to express the result in a smaller unit like centimeters. To convert the wavelength from meters to centimeters, we need to multiply it by 100 (since there are 100 centimeters in a meter). λ = 0.3529 m * 100 cm/m = 35.29 cm The wavelength of the electromagnetic waves used for cell phone communications in the 850 MHz band is approximately 35.29 centimeters.

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 Vaia!

Key Concepts

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

Wavelength Calculation
The concept of wavelength is fundamental to understanding how electromagnetic waves operate. Wavelength describes the distance between consecutive peaks of a wave. In this context, it is related to how cell phone communications work. Exposure to the concept helps in grasping how waves propagate through space. To calculate wavelength, we need to know two things:
  • The speed at which the wave travels (for electromagnetic waves like light, this is constant).
  • The frequency of the wave, which refers to how often the wave peaks in a certain period.
The relationship between these variables is given by the formula:\[ \lambda = \frac{c}{f} \]Where:- \(\lambda\) (lambda) stands for wavelength,- \(c\) is the speed of light,- \(f\) is the frequency.By plugging in the given frequency and the speed of light, we can determine the wavelength of the electromagnetic wave in question. In practice, this allows us to convert our understanding from how often a wave cycles per second (frequency) into a tangible distance measurement (wavelength), enhancing our comprehension of the wave's characteristics.
Frequency Conversion
Frequency conversion is a necessary step in calculating the wavelength of electromagnetic waves. Initially, the frequency might be given in megahertz (MHz), a common unit in telecommunications. Converting this into hertz (Hz), the SI unit for frequency, ensures a uniform and standardized approach.To convert MHz to Hz:- Remember that 1 MHz is equivalent to 1,000,000 Hz. - Thus, you need to multiply the frequency value in MHz by \(10^6\) to convert it to Hz.For instance, an 850 MHz frequency becomes 850 * \(10^6\) Hz, equating to \(8.50 \times 10^8\) Hz. Using Hz in further calculations makes computations with variables like the speed of light more straightforward and consistent. Successful frequency conversion is crucial as it seamlessly integrates with other physical laws and equations, making your scientific inquiries accurate and effective.
Speed of Light
The speed of light is a constant and essential part of the wavelength calculation formula. It describes how fast light and all other electromagnetic waves travel through a vacuum. The speed is approximately \(3.00 \times 10^8\) meters per second (m/s).This constant is instrumental in determining the characteristics of electromagnetic waves. For example, in the wavelength formula \(\lambda = \frac{c}{f}\), the speed of light \(c\) provides a direct relationship between how often waves oscillate (frequency) and the physical distance between wave peaks (wavelength).Furthermore, it's crucial to understand that:
  • The speed of light remains unaltered in a vacuum but can change when waves enter different mediums, such as glass or water.
  • This speed forms a foundation not only for wavelength calculations but also for broader fields like relativity and quantum mechanics.
Having a firm grasp on the speed of light helps in multiple scientific domains, offering clarity and consistency in complex calculations and theories.

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

Calculate the average value of the Poynting vector, \(S_{\text {ave }}\) for an electromagnetic wave having an electric field of amplitude \(100 . \mathrm{V} / \mathrm{m}\) a) What is the average energy density of this wave? b) How large is the amplitude of the magnetic field?

Determine the distance in feet that light can travel in vacuum during \(1.00 \mathrm{~ns}\).

Quantum theory says that electromagnetic waves actually consist of discrete packets-photons-each with energy \(E=\hbar \omega,\) where \(\hbar=1.054573 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's reduced constant and \(\omega\) is the angular frequency of the wave. a) Find the momentum of a photon. b) Find the angular momentum of a photon. Photons are circularly polarized; that is, they are described by a superposition of two plane-polarized waves with equal field amplitudes, equal frequencies, and perpendicular polarizations, one-quarter of a cycle \(\left(90^{\circ}\right.\) or \(\pi / 2\) rad \()\) out of phase, so the electric and magnetic field vectors at any fixed point rotate in a circle with the angular frequency of the waves. It can be shown that a circularly polarized wave of energy \(U\) and angular frequency \(\omega\) has an angular momentum of magnitude \(L=U / \omega .\) (The direction of the angular momentum is given by the thumb of the right hand, when the fingers are curled in the direction in which the field vectors circulate. c) The ratio of the angular momentum of a particle to \(\hbar\) is its spin quantum number. Determine the spin quantum number of the photon.

A \(5.00-\mathrm{mW}\) laser pointer has a beam diameter of \(2.00 \mathrm{~mm}\) a) What is the root-mean-square value of the electric field in this laser beam? b) Calculate the total electromagnetic energy in \(1.00 \mathrm{~m}\) of this laser beam.

The voltage across a cylindrical conductor of radius \(r\), length \(L\), and resistance \(R\) varies with time. The timevarying voltage causes a time-varying current, \(i\), to flow in the cylinder. Show that the displacement current equals \(\epsilon_{0} \rho d i / d t,\) where \(\rho\) is the resistivity of the conductor.

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.

Sign-up for free