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(a) In \(\mathrm{MeV} / c\), what is the magnitude of the momentum associated with a photon having an energy equal to the electron rest energy? What are the (b) wavelength and (c) frequency of the corresponding radiation?

Short Answer

Expert verified
Momentum: 0.511 MeV/c, Wavelength: 2.43 pm, Frequency: 1.24 脳 10虏鈦 Hz.

Step by step solution

01

Understand the Problem

We are looking to find the momentum, wavelength, and frequency of a photon whose energy equals the rest energy of an electron. We know the electron rest energy is given by the formula: \( E = m_e c^2 \), where \( m_e \approx 0.511 \, \text{MeV}/c^2 \) is the electron mass.
02

Calculate the Photon Energy

Since the photon's energy equals the electron rest energy, we have \( E = 0.511 \, \text{MeV} \). This will be used in the subsequent calculations.
03

Calculate the Photon Momentum

For a photon, the momentum \( p \) is related to its energy by the formula: \( E = pc \), where \( c \) is the speed of light. Therefore, the momentum of the photon is given by \( p = \frac{E}{c} \). Substituting \( E = 0.511 \, \text{MeV} \), we obtain \( p = 0.511 \, \text{MeV}/c \).
04

Calculate the Wavelength

The wavelength \( \lambda \) is related to the momentum by the de Broglie wavelength formula: \( \lambda = \frac{h}{p} \), where \( h = 4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \) is Planck's constant. Converting Planck's constant to \( \text{MeV} \cdot \text{s} \): \( h = 4.135667696 \times 10^{-21} \, \text{MeV} \cdot \text{s} \). Therefore, \( \lambda = \frac{4.135667696 \times 10^{-21} \, \text{MeV} \cdot \text{s}}{0.511 \, \text{MeV}/c} \), simplifying, we get \( \lambda \approx 2.43 \times 10^{-12} \text{m} \).
05

Calculate the Frequency

The frequency \( f \) is related to the energy by \( E = hf \). Rearranging gives \( f = \frac{E}{h} \). Substituting \( E = 0.511 \, \text{MeV} \) and converting \( E \) to \( 511 \times 10^3 \, \text{eV} \): \( f = \frac{511 \times 10^3 \, \text{eV}}{4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s}} \approx 1.24 \times 10^{20} \text{Hz} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Rest Energy
Every electron has intrinsic energy, known as rest energy, even when it isn't moving. This energy can be calculated using Einstein's famous equation: \[ E = m_e c^2 \] where \( E \) is the rest energy, \( m_e \) is the mass of the electron, approximately \( 0.511 \, \text{MeV}/c^2 \), and \( c \) is the speed of light. This rest energy is a fundamental concept of physics as it establishes a basis for comparing with other forms of energy, like photon energy. Since photons are particles of light with energy, they can be directly compared to the electron rest energy. Understanding this concept allows us to analyze situations in fields such as, quantum physics and particle physics, where particle interactions are considered.
Photon Momentum
Photons, despite having no mass, do possess momentum, which is derived from their energy. The momentum \( p \) of a photon can be calculated using the equation: \[ E = pc \] where \( E \) is the energy of the photon and \( c \) is the speed of light. Re-arranging this equation gives us:\[ p = \frac{E}{c} \] Given that the energy of the photon is equal to the electron's rest energy, the photon's momentum will be \( 0.511 \, \text{MeV}/c \). This demonstrates how momentum can be a characteristic of massless particles. Photon momentum plays a crucial role in understanding light's interaction with matter, leading to practical applications from solar sails to laser technology.
Wavelength Calculation
The wavelength of a photon is inversely related to its momentum. We use the de Broglie wavelength equation: \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the wavelength, \( h \) is Planck's constant \( \approx 4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \), and \( p \) is the photon's momentum. First, we convert Planck's constant to \( \text{MeV} \cdot \text{s} \): \( h = 4.135667696 \times 10^{-21} \, \text{MeV} \cdot \text{s} \). Substituting \( p = 0.511 \, \text{MeV}/c \), we solve for \( \lambda \), yielding: \[ \lambda \approx 2.43 \times 10^{-12} \text{m} \] This calculation is essential in fields like spectroscopy and diffraction studies, where the wavelength determines the type of electromagnetic radiation.
Frequency Calculation
Frequency is a measure of how often oscillations occur in a wave per unit time. The energy-frequency relationship for a photon is given by the equation: \[ E = hf \] where \( E \) is the energy, \( h \) is Planck's constant, and \( f \) is the frequency. Rearranging, we find:\[ f = \frac{E}{h} \] Substituting \( E = 0.511 \, \text{MeV} \) (converted to \( 511 \times 10^3 \, \text{eV} \)) and \( h = 4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s} \), we calculate:\[ f \approx 1.24 \times 10^{20} \text{Hz} \] The frequency of a photon is fundamental to understanding phenomena such as the electromagnetic spectrum, where different frequencies correspond to different types of electromagnetic radiation like visible light, X-rays, or microwaves.

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