Problem 52
Question
A photon has an energy of \(2.93 \times 10^{-25} \mathrm{J} .\) What is its frequency? What type of electromagnetic radiation is the photon?
Step-by-Step Solution
Verified Answer
The frequency is approximately \(4.42 \times 10^8 \text{ Hz}\), classifying it as a radio wave.
1Step 1: Identify Relevant Formula
To find the frequency of a photon, we use the formula that relates energy and frequency: \[ E = h \cdot f \]where:- \( E \) is the energy of the photon (in joules),- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \text{ J} \cdot \text{s} \)),- \( f \) is the frequency of the photon (in hertz).
2Step 2: Rearrange Formula for Frequency
Rearrange the formula to solve for frequency \( f \): \[ f = \frac{E}{h} \] This will allow us to insert the given values for \( E \) and \( h \) to calculate \( f \).
3Step 3: Insert Values and Solve
Substitute \( E = 2.93 \times 10^{-25} \text{ J} \) and \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \) into the formula: \[ f = \frac{2.93 \times 10^{-25}}{6.626 \times 10^{-34}} \] Perform the division to find the frequency of the photon.
4Step 4: Calculate Frequency
Perform the calculation:\[ f = \frac{2.93 \times 10^{-25}}{6.626 \times 10^{-34}} \approx 4.42 \times 10^8 \text{ Hz} \]This is the frequency of the photon.
5Step 5: Determine Type of Radiation
Electromagnetic radiation types are categorized by frequency. A frequency of \( 4.42 \times 10^8 \text{ Hz} \) falls within the radio wave portion of the electromagnetic spectrum.
Key Concepts
Frequency CalculationPlanck's ConstantElectromagnetic SpectrumRadio Waves
Frequency Calculation
Calculating the frequency of a photon is essential when understanding various electromagnetic phenomena. Frequency (\( f \)) represents the number of wave cycles that pass through a point per second and is measured in Hertz (\( \text{Hz} \)).
To connect frequency and energy, we use the formula: \[ E = h \times f \]Here, \( E \) is the energy of the photon and \( h \) is Planck's constant.
To solve for frequency, you can rearrange the formula to: \[ f = \frac{E}{h} \]
For instance, if energy is given as \( 2.93 \times 10^{-25} \text{ J} \) and Planck's constant is \( 6.626 \times 10^{-34} \text{ J} \cdot \text{ s} \), the frequency would be: \[ f = \frac{2.93 \times 10^{-25}}{6.626 \times 10^{-34}} \approx 4.42 \times 10^8 \text{ Hz} \]This calculation reveals how often the wave repeats within one second.
To connect frequency and energy, we use the formula: \[ E = h \times f \]Here, \( E \) is the energy of the photon and \( h \) is Planck's constant.
To solve for frequency, you can rearrange the formula to: \[ f = \frac{E}{h} \]
For instance, if energy is given as \( 2.93 \times 10^{-25} \text{ J} \) and Planck's constant is \( 6.626 \times 10^{-34} \text{ J} \cdot \text{ s} \), the frequency would be: \[ f = \frac{2.93 \times 10^{-25}}{6.626 \times 10^{-34}} \approx 4.42 \times 10^8 \text{ Hz} \]This calculation reveals how often the wave repeats within one second.
Planck's Constant
Planck's constant is a cornerstone of quantum mechanics. It bridges the gap between the energy of a photon and its frequency.
This constant (\( h \)) is approximately \( 6.626 \times 10^{-34} \text{ J} \cdot \text{ s} \).
While it might appear minuscule, it's extremely significant in the realm of quantum physics.
This constant (\( h \)) is approximately \( 6.626 \times 10^{-34} \text{ J} \cdot \text{ s} \).
While it might appear minuscule, it's extremely significant in the realm of quantum physics.
- It expresses the quantized nature of energy, showing that energy comes in specific packets or quanta.
- Planck's constant is essential for deriving equations that explain phenomena on atomic and subatomic scales.
By using Planck's constant, one can convert between the energy of a photon and its observable frequency.
Electromagnetic Spectrum
The electromagnetic spectrum is a range of all possible frequencies of electromagnetic radiation.
Different types of electromagnetic waves exist, each having unique properties and uses.
These types are separated based on their frequency and wavelength.
The placement of a wave on the spectrum impacts its behavior and interaction with matter.
For instance, waves with lower frequencies like radio waves can pass through walls, while higher frequencies like X-rays cannot.
Different types of electromagnetic waves exist, each having unique properties and uses.
These types are separated based on their frequency and wavelength.
- Radio waves
- Microwaves
- Infrared rays
- Visible light
- Ultraviolet rays
- X-rays
- Gamma rays
The placement of a wave on the spectrum impacts its behavior and interaction with matter.
For instance, waves with lower frequencies like radio waves can pass through walls, while higher frequencies like X-rays cannot.
Radio Waves
Radio waves are a type of electromagnetic radiation that possess the longest wavelength and the lowest frequency on the electromagnetic spectrum.
Due to their low energy, they are highly versatile and are used in numerous applications. Considered safe for our usual exposure, they are integral to our daily technology.
Due to their low energy, they are highly versatile and are used in numerous applications. Considered safe for our usual exposure, they are integral to our daily technology.
- Communication: Radio and TV broadcasting, wireless communication, and satellite transmissions.
- Navigation: GPS signals rely heavily on radio waves.
- Exploration: Radio waves are used in radar systems to determine the speed and location of objects.
Other exercises in this chapter
Problem 49
What is the energy of a photon of red light that has a frequency of \(4.48 \times 10^{14} \mathrm{Hz}\) ?
View solution Problem 51
What is the energy of an ultraviolet photon that has a wavelength of \(1.18 \times 10^{-8} \mathrm{m} ?\)
View solution Problem 53
A photon has an energy of \(1.10 \times 10^{-13} \mathrm{J.}\) What is the photon's wavelength? What type of electromagnetic radiation is it?
View solution Problem 54
Spacecraft How long does it take a radio signal from the Voyager spacecraft to reach Earth if the distance between Voyager and Earth is \(2.72 \times 10^{9} \ma
View solution