Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 2134

The frequency of electromagnetic wave having wavelength \(25 \mathrm{~mm}\) is \(\quad \mathrm{Hz}\) (A) \(1.2 \times \overline{10^{10}}\) (B) \(7.5 \times 10^{5}\) (C) \(1.2 \times 10^{8}\) (D) \(7.5 \times 10^{6}\)

Short Answer

Expert verified
The frequency of an electromagnetic wave having wavelength \(25~mm\) is \(1.2 \times 10^{10}~Hz\). The correct answer is (A).

Step by step solution

01

Write the given values

We are given the wavelength \(\lambda = 25~mm\) or \(2.5 \times 10^{-2}~m\) (remember to convert millimeters to meters). The speed of light \(c = 3 \times 10^8 m/s\).
02

Use the formula to find the frequency

We will now use the formula \(f = \frac{c}{\lambda}\) to find the frequency: \(f = \frac{3 \times 10^8~m/s}{2.5 \times 10^{-2}~m}\)
03

Perform the calculation

Now we will proceed to perform the calculation: \(f = \frac{3 \times 10^8}{2.5 \times 10^{-2}}\) \(f = \frac{3}{2.5} \times \frac{10^8}{10^{-2}}\) \(f = 1.2 \times 10^{10}~Hz\)
04

Identify the correct option

Comparing our obtained frequency, \(f = 1.2 \times 10^{10}~Hz\), to the given options, the correct answer is: (A) \(1.2 \times 10^{10}~Hz\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Electromagnetic waves travelling in a medium which has relative permeability \(1.3\) and relative permittivity \(2.14\) speed of electromagnetic waves in this medium will be (A) \(3.6 \times 10^{8} \mathrm{~m} / \mathrm{s}\) (B) \(1.8 \times 10^{8} \mathrm{~m} / \mathrm{s}\) (C) \(1.8 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (D) \(13.6 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

In a plane electromagnetic wave, the electric field oscillates sinusoidaly at a frequency of \(2.0 \times 10^{10} \mathrm{~Hz}\). if the peak value of electric field is \(60 \mathrm{Vm}^{-1}\) the average energy density (in \(\mathrm{Jm}^{-3}\) ) of the magnetic field of the wave will be (given \(\left.\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}\right)\) (A) \(2 \pi \times 10^{-7}\) (B) \((1 / 2 \pi) \times 10^{-7}\) (C) \(4 \pi \times 10^{-7}\) (D) \((1 / 4 \pi) \times 10^{-7}\)

Infrared radiations are detected by (A) spectrometer (B) bolometer (C) photocells (D) geiger tubes

The frequency of light wave of wavelength \(5000 \mathrm{~A}\) is \(\mathrm{Hz}\) (A) \(6 \times 10^{14}\) (B) \(1.5 \times 10^{-2}\) (C) \(1.5\) (D) \(6 \times 10^{1}\)

If \(\mathrm{V}_{\mathrm{r}}, \mathrm{V}_{\mathrm{x}}\) and \(\mathrm{V}_{\mathrm{m}}\) are the velocity of the \(\gamma\) rays, \(\mathrm{x}\) rays, micro waves respectively in space, then (A) \(\mathrm{V}_{\gamma}<\mathrm{V}_{\mathrm{x}}<\mathrm{V}_{\mathrm{m}}\) (B) \(\mathrm{V}_{\mathrm{r}}=\mathrm{V}_{\mathrm{x}}=\mathrm{V}_{\mathrm{m}}\) (C) \(\mathrm{V}_{\mathrm{r}}^{\prime}>\mathrm{V}_{\mathrm{x}}>\mathrm{V}_{\mathrm{m}}\) (D) \(\mathrm{V}_{\mathrm{r}}>\mathrm{V}_{\mathrm{x}}<\mathrm{V}_{\mathrm{m}}\)
See all solutions

Recommended explanations on English Textbooks

Discourse

Read Explanation

Blog

Read Explanation

Single Paragraph Essay

Read Explanation

Essay Plan

Read Explanation

Syntax

Read Explanation

Synthesis Essay

Read Explanation
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