Problem 12
Question
$$ \text { Find the frequency of an electromagnetic wave if its wavelength is } 3.25 \times 10^{-8} \mathrm{~m} \text { . } $$
Step-by-Step Solution
Verified Answer
The frequency of the electromagnetic wave is \( 9.23 \times 10^{15} \text{ Hz} \).
1Step 1: Recall the Speed of Light Formula
Electromagnetic waves, including light, travel at a constant speed in a vacuum. The speed of light \( c \) is approximately \( 3.00 \times 10^8 \text{ m/s} \). The formula that relates speed, frequency (\( f \)), and wavelength (\( \lambda \)) is \( c = f \times \lambda \).
2Step 2: Rearrange the Formula to Solve for Frequency
Since we are looking for the frequency, we need to rearrange the formula to isolate \( f \). This gives us \( f = \frac{c}{\lambda} \).
3Step 3: Plug in the Known Values
Now, substitute the known values into the equation. The speed of light \( c \) is \( 3.00 \times 10^8 \text{ m/s} \) and the wavelength \( \lambda \) is given as \( 3.25 \times 10^{-8} \text{ m} \).
4Step 4: Carry Out the Division
Calculate the frequency using the rearranged equation: \[ f = \frac{3.00 \times 10^8}{3.25 \times 10^{-8}} \] Divide the numbers: \[ f = 3.00 \div 3.25 \times 10^{8 - (-8)} \] \[ f \approx 0.923 \times 10^{16} \]
5Step 5: Adjust the Scientific Notation
Change the result from Step 4 to proper scientific notation: \[ f \approx 9.23 \times 10^{15} \text{ Hz} \] This is the frequency of the electromagnetic wave.
Key Concepts
Frequency CalculationWavelengthSpeed of LightScientific Notation
Frequency Calculation
To determine the frequency of an electromagnetic wave, one of the fundamental relationships we use connects the speed of light, frequency, and wavelength. The formula is:
- \( c = f \times \lambda \)
- \( f = \frac{c}{\lambda} \)
Wavelength
Wavelength is a key factor in determining the properties of an electromagnetic wave. It is the distance between consecutive peaks of a wave. Wavelength is denoted by the Greek letter \( \lambda \) and is usually measured in meters (m). In the context of electromagnetic waves, wavelength helps determine the wave's energy and color (for visible light). It is inversely proportional to frequency, meaning as wavelength increases, frequency decreases and vice versa. This is because:
- \( f = \frac{c}{\lambda} \)
Speed of Light
The speed of light, denoted as \( c \), is a constant in a vacuum that measures how fast light travels. Its approximate value is \( 3.00 \times 10^{8} \text{ m/s} \). This constant is essential for many calculations in physics, especially in the study of electromagnetic waves. The idea that light travels at this constant speed across any vacuum was pivotal in the development of Einstein's theory of relativity.
The speed of light allows us to connect frequency and wavelength through its equation:
The speed of light allows us to connect frequency and wavelength through its equation:
- \( c = f \times \lambda \)
Scientific Notation
Scientific notation is a way to express very large or very small numbers in a compact form. It's a crucial tool in fields like physics, where extreme values are common. A number in scientific notation is split into two components: a coefficient and a power of ten. For example, \( 3.00 \times 10^{8} \) represents the speed of light. Here, 3.00 is the coefficient, and \( 10^{8} \) indicates how many times the coefficient should be multiplied by ten.
This method simplifies calculations, as the powers of ten can be easily added or subtracted. For division, the exponents are subtracted: when \( c \), \( f \), and \( \lambda \) are expressed in scientific notation, determining frequency through:
This method simplifies calculations, as the powers of ten can be easily added or subtracted. For division, the exponents are subtracted: when \( c \), \( f \), and \( \lambda \) are expressed in scientific notation, determining frequency through:
- \( f = \frac{3.00 \times 10^{8}}{3.25 \times 10^{-8}} \)
Other exercises in this chapter
Problem 12
Find the intensity of a light source that produces illumination of \(5.50 \mathrm{ft}-\) candles at \(9.85 \mathrm{ft}\) from the source.
View solution Problem 12
An AM radio signal has a frequency of \(65 \overline{0} \mathrm{kHz}\). What is the energy of a photon of that electromagnetic radiation?
View solution Problem 13
(a) How long does it take for light to reach the earth from Mars when the separation of the two planets is at its smallest? The earth's orbital radius is 143 mi
View solution Problem 14
Find the intensity of a light source that produces an illumination of \(5.28\) lux at \(6.50 \mathrm{~m}\) from the source.
View solution