Problem 95
Question
What is the wavelength of light with a frequency of \(5.77 \times 10^{14} \mathrm{Hz} ?\)
Step-by-Step Solution
Verified Answer
The wavelength is approximately 520 nm.
1Step 1: Understand the Formula
The relationship between the speed of light \( c \), frequency \( f \), and wavelength \( \lambda \) is given by the equation \( c = \lambda \times f \). The speed of light \( c \) is approximately \( 3 \times 10^8 \text{ m/s} \).
2Step 2: Rearrange the Formula
To find the wavelength \( \lambda \), rearrange the equation \( c = \lambda \times f \) to solve for \( \lambda \). The rearranged formula is \( \lambda = \frac{c}{f} \).
3Step 3: Substitute Values into the Equation
Using the rearranged formula, substitute the known values: \( c = 3 \times 10^8 \text{ m/s} \) and \( f = 5.77 \times 10^{14} \text{ Hz} \). This gives us \( \lambda = \frac{3 \times 10^8}{5.77 \times 10^{14}} \).
4Step 4: Perform the Calculation
Calculate the wavelength \( \lambda \) by dividing the speed of light by the frequency. This calculates to \( \lambda \approx 5.20 \times 10^{-7} \text{ m} \).
5Step 5: Express the Wavelength in Nanometers
To express the wavelength in nanometers (nm), recall that \(1 \text{ m} = 10^9 \text{ nm} \). Therefore, \( \lambda \approx 5.20 \times 10^{-7} \text{ m} = 520 \text{ nm} \).
Key Concepts
FrequencySpeed of LightNanometersPhysics Equations
Frequency
Frequency is the number of oscillations or cycles that occur in a wave per unit of time. Think of frequency as how often something repeats itself over a set period.
It's usually measured in Hertz (Hz), which represents cycles per second. For example, if a light wave has a frequency of 5.77 x 10^{14} Hz, it means that the wave completes 5.77 x 10^{14} cycles every second.
Frequency is a fundamental concept in physics, especially in the study of waves. It affects other properties of waves such as wavelength. By knowing the frequency of a wave, you can learn more about how the wave behaves.
It's usually measured in Hertz (Hz), which represents cycles per second. For example, if a light wave has a frequency of 5.77 x 10^{14} Hz, it means that the wave completes 5.77 x 10^{14} cycles every second.
Frequency is a fundamental concept in physics, especially in the study of waves. It affects other properties of waves such as wavelength. By knowing the frequency of a wave, you can learn more about how the wave behaves.
Speed of Light
The speed of light is a fundamental constant in physics, typically denoted by the symbol \( c \).
In a vacuum, light travels at an incredible speed of approximately \( 3 \times 10^8 \text{ meters per second} \). It's one of the fastest speeds known in the universe and serves as a speed limit for information and matter.
The speed of light is crucial to understanding various physical phenomena, from light refraction to relativity. It also plays a key role in equations involving light, such as finding the wavelength when the frequency is known, using the formula \( c = \lambda \times f \).
In a vacuum, light travels at an incredible speed of approximately \( 3 \times 10^8 \text{ meters per second} \). It's one of the fastest speeds known in the universe and serves as a speed limit for information and matter.
The speed of light is crucial to understanding various physical phenomena, from light refraction to relativity. It also plays a key role in equations involving light, such as finding the wavelength when the frequency is known, using the formula \( c = \lambda \times f \).
Nanometers
Nanometers are a unit of length in the metric system, used to measure things at the microscopic scale.
One nanometer (nm) is one billionth of a meter, or \( 1 ext{ nm} = 10^{-9} ext{ m} \). Because light wavelengths are generally very small, they are often expressed in nanometers.
For example, visible light has wavelengths ranging from about 400 nm (violet) to 700 nm (red). In this scenario, when the wavelength calculated was \( 5.20 \times 10^{-7} \text{ m} \), converting it to nanometers gives us 520 nm, bringing it comfortably into the range of visible light.
One nanometer (nm) is one billionth of a meter, or \( 1 ext{ nm} = 10^{-9} ext{ m} \). Because light wavelengths are generally very small, they are often expressed in nanometers.
For example, visible light has wavelengths ranging from about 400 nm (violet) to 700 nm (red). In this scenario, when the wavelength calculated was \( 5.20 \times 10^{-7} \text{ m} \), converting it to nanometers gives us 520 nm, bringing it comfortably into the range of visible light.
Physics Equations
Physics equations are mathematical expressions that describe physical laws and relationships between different quantities.
In the context of wave equations, an important relationship explains how frequency, wavelength, and the speed of light interact with each other: \( c = \lambda \times f \).
By rearranging this equation, you can solve for any of the three variables if the other two are known. For example, if you know the speed of light and the frequency, you can find the wavelength using \( \lambda = \frac{c}{f} \).
Equations like these are tools that physicists use to predict and understand how different forces and objects behave in the universe, making them invaluable for anyone studying physics.
In the context of wave equations, an important relationship explains how frequency, wavelength, and the speed of light interact with each other: \( c = \lambda \times f \).
By rearranging this equation, you can solve for any of the three variables if the other two are known. For example, if you know the speed of light and the frequency, you can find the wavelength using \( \lambda = \frac{c}{f} \).
Equations like these are tools that physicists use to predict and understand how different forces and objects behave in the universe, making them invaluable for anyone studying physics.
Other exercises in this chapter
Problem 93
For an atom of tin in the ground state, write the electron configuration using noble-gas notation, and draw its electron-dot structure.
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What is the maximum number of electrons that can be contained in an atom's orbitals having the following principal quantum numbers? $$\begin{array}{ll}{\text {
View solution Problem 97
How many orientations are possible for the orbitals related to each sublevels? $$\begin{array}{ll}{\text { a. } \mathrm{s}} & {\text { c. d }} \\ {\text { b. }
View solution Problem 98
Which elements have only two electrons in their electron- dot structures: hydrogen, helium, lithium, aluminum, calcium, cobalt, bromine, krypton, or barium?
View solution