Problem 8
Question
$$ \begin{aligned} &\text { Find the energy of a photon of electromagnetic radiation with frequency }\\\ &9.20 \times 10^{16} \mathrm{~Hz} \end{aligned} $$
Step-by-Step Solution
Verified Answer
The energy of the photon is \(6.10 \times 10^{-17} \text{ J}\).
1Step 1: Understand the Formula
The energy \(E\) of a photon can be calculated using the formula: \[ E = h \cdot f \]where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ Js}\) and \(f\) is the frequency of the radiation.
2Step 2: Substitute Known Values
Plug the given frequency and Planck's constant into the formula. \[ E = 6.626 \times 10^{-34} \text{ Js} \times 9.20 \times 10^{16} \text{ Hz} \]
3Step 3: Calculate the Energy
Multiply the constants and the frequency:\[ E = 6.626 \times 10^{-34} \times 9.20 \times 10^{16} = 6.09792 \times 10^{-17} \text{ J} \]
Key Concepts
Planck's ConstantElectromagnetic RadiationFrequency of Radiation
Planck's Constant
Planck's constant is a fundamental quantity in the quantum world. Denoted by the letter \( h \), it is crucial in the formula to calculate the energy of a photon. Planck's constant has a value of \( 6.626 \times 10^{-34} \text{ Js} \). Joule-seconds is the unit of this constant as it refers to energy (in joules) times time (in seconds).
This constant is significant for understanding the quantized nature of energy transfer in the realm of quantum mechanics. In simple terms, it tells us that energy changes in steps defined by this constant rather than in a continuous flow.
This constant is significant for understanding the quantized nature of energy transfer in the realm of quantum mechanics. In simple terms, it tells us that energy changes in steps defined by this constant rather than in a continuous flow.
- Planck's constant connects the energy of particles like photons with their frequency.
- It's a foundational pillar of physics theories like quantum mechanics.
Electromagnetic Radiation
Electromagnetic radiation is a form of energy that travels through space at the speed of light. It is composed of oscillating electric and magnetic fields that propagate through space. This type of radiation includes visible light as well as ultraviolet, infrared, X-rays, microwaves, and radio waves.
These waves are classified by their wavelengths or frequencies, leading to the electromagnetic spectrum. The energy carried by this radiation depends on the frequency of the waves. Higher frequency waves, like X-rays, carry more energy than lower frequency waves like radio waves.
These waves are classified by their wavelengths or frequencies, leading to the electromagnetic spectrum. The energy carried by this radiation depends on the frequency of the waves. Higher frequency waves, like X-rays, carry more energy than lower frequency waves like radio waves.
- This radiation can behave both as waves and particles, known as wave-particle duality.
- Photons are the particle components of electromagnetic radiation.
Frequency of Radiation
The frequency of radiation is a measure of how many times a wave cycle occurs within a second. It's measured in hertz (Hz), which means cycles per second.
This concept is essential for defining the energy of photons. According to the relationship \( E = h \cdot f \), the energy \( E \) of a photon is directly proportional to its frequency \( f \). Hence, a higher frequency means more energy.
This concept is essential for defining the energy of photons. According to the relationship \( E = h \cdot f \), the energy \( E \) of a photon is directly proportional to its frequency \( f \). Hence, a higher frequency means more energy.
- Frequency directly influences the energy that the electromagnetic radiation delivers.
- It is inversely related to wavelength, meaning higher frequency radiation has shorter wavelengths.
Other exercises in this chapter
Problem 6
$$ \begin{aligned} &c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} \\ &f=415 \mathrm{~Hz} \\ &\lambda=? \end{aligned} $$
View solution Problem 7
$$ \text { Find the frequency of electromagnetic radiation with energy } 3.65 \times 10^{-23} \mathrm{~J} . $$
View solution Problem 8
$$ \begin{aligned} &c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} \\ &\lambda=9.23 \mathrm{~km} \\ &f=? \end{aligned} $$
View solution Problem 9
Find the wavelength of a radio wave from an AM station broadcasting at a frequency of \(14 \overline{0} 0 \mathrm{kHz}\).
View solution