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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) \(1.2 \times 10^{10}~Hz\)

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Most popular questions from this chapter

The amplitude of the magnetic field part of an electromagnetic wave in vacuum is \(\mathrm{Bm}=510 \mathrm{nT}\). Then the amplitude of the electric part of the wave is (A) \(1.53 \times 10^{11} \mathrm{~V} / \mathrm{m}\) (B) \(1.53 \mathrm{~V} / \mathrm{m}\) (C) \(1.53 \times 10^{2} \mathrm{~V} / \mathrm{m}\) (D) \(1.53 \times 10^{8} \mathrm{~V} / \mathrm{m}\)

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