Logout Open In App
Open In App
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

Chapter 15: Problem 2166

The frequency \(1057 \mathrm{MHz}\) of radiation arising from two close energy levels in hydrogen belongs to (A) radio waves (B) infrared waves (C) micro waves (D) \gamma rays

Short Answer

Expert verified
The given frequency is \(1057 MHz\), which equals \(1057 \times 10^6\) Hz. Comparing this with the frequency ranges of different electromagnetic waves, it falls within the range of radio waves (< \(10^9\) Hz). Therefore, the radiation belongs to (A) radio waves.

Step by step solution

01

Convert the frequency to Hz

Given the frequency is 1057 MHz, convert megahertz (MHz) to hertz (Hz) by multiplying by \(10^6\): Frequency = \(1057 \times 10^6\) Hz
02

Compare the frequency with the frequency ranges of electromagnetic waves

We will now compare the given frequency of \(1057 \times 10^6\) Hz to the frequency ranges of the different types of electromagnetic waves. - Radio waves: < \(10^9\) Hz - Infrared waves: \(10^{12}\space \textrm{to}\space 4 \times 10^{14}\) Hz - Micro waves: \(10^9 \space \textrm{to}\space 10^{12}\) Hz - Gamma rays: > \(10^{19}\) Hz
03

Determine the category of the electromagnetic wave

Since the given frequency is \(1057 \times 10^6\) Hz, which is less than \(10^9\) Hz, it falls within the frequency range of radio waves. So, the correct answer is (A) radio waves.

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

According to Maxwell, a changing electric field produces (A) emf (B) Electric current (C) magnetic field (D) radiation pressure

An electromagnetic wave going through vacuum is described by $E=E_{0} \sin (k x-\cot )$. Which of the following is independent of the wavelength? (A) \(\omega\) (B) \((\mathrm{k} / \mathrm{c})\) (C) \(\mathrm{k}_{\mathfrak{e}}\) (D) \(\mathrm{k}\)

The maximum value of \(\mathrm{E}^{-}\) in an electromagnetic waves in air is equal to \(6.0 \times 10^{-4} \mathrm{Vm}^{-1}\). The maximum value of \(\mathrm{B}^{-}\) is (A) \(1.8 \times 10^{5} \mathrm{~T}\) (B) \(2.0 \times 10^{4} \mathrm{~T}\) (C) \(2.0 \times 10^{-12} \mathrm{~T}\) (D) \(1.8 \times 10^{13} \mathrm{~T}\)

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) \(6 \times 10^{1}\)

The rms value of the electric field of the light coming from the sun is $720 \mathrm{~N} / \mathrm{c}$. The average total energy density of the electromagnetic wave is (A) \(4.58 \times 10^{-6} \mathrm{Jm}^{-3}\) (B) \(6.3 \times 10^{-9} \mathrm{Jm}^{-3}\) (C) \(81.35 \times 10^{-12} \mathrm{Jm}^{-3}\) (D) \(3.3 \times 10^{-3} \mathrm{Jm}^{-3}\)
See all solutions

Recommended explanations on English Textbooks

Phonology

Read Explanation

Language Acquisition

Read Explanation

Language Analysis

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Language and Social Groups

Read Explanation

Text Comparison

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