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Magazine Chapter 3: Problem 42
A photon of wavelength \(6.13 \mathrm{pm}\) scatters from a free electron at rest. After the interaction, the electron is observed to be moving in the direction of the original photon. Find the momentum of the electron.
Short Answer
Expert verified
The momentum of the electron is approximately \(1.08 \times 10^{-22} \mathrm{kg \cdot m/s} \).
Step by step solution
01
- Define the Compton wavelength shift formula
The Compton wavelength shift formula is given by \[ \Delta \lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos \theta) \] Where: \[ \lambda \] - initial wavelength of the photon, \[ \lambda' \] - scattered wavelength of the photon, \[ h \] - Planck鈥檚 constant, \[ m \] - mass of the electron, \[ c \] - speed of light, \[ \theta \] - is the scattering angle of the photon.
02
- Set up the known values
Given: The initial wavelength of the photon is \[ 6.13 \mathrm{pm} \] Since the photon scatters and moves in the original direction, \[ \theta = 0掳 \] The mass of an electron \[ m = 9.11 \times 10^{-31} \mathrm{kg} \] and Planck's constant \[ h = 6.626 \times 10^{-34} \mathrm{Js} \].
03
- Calculate Compton Shift
Using the Compton wavelength shift formula for \[ \theta = 0 \], We get: \[ \Delta \lambda = \frac{6.626 \times 10^{-34} }{9.11 \times 10^{-31} \times 3 \times 10^8} (1 - \cos (0)) \] Since \[ \cos(0掳) = 1 \], \[ \Delta \lambda = 0 \] There is no wavelength change in this case because the angle is 0.
04
- Use conservation of momentum
Since the scattered wavelength does not change, energy and momentum can be treated using initial wavelength: The energy, \[ \mathrm{E} \] of the photon is \[ E = \frac{hc}{\lambda} \] For \[ \lambda = 6.13 \times 10^{-12}\mathrm{m} \], We get: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6.13 \times 10^{-12}} \approx 3.24 \times 10^{-14} \mathrm{J} \]
05
- Calculate the momentum of the electron
The momentum \[ \mathrm{p} \] is given by \[ p = \frac{E}{c} \]. Using \[ E = 3.24 \times 10^{-14} \mathrm{J} \] and \[ c = 3 \times 10^8 \mathrm{m/s} \], We find: \[ p = \frac{3.24 \times 10^{-14}}{3 \times 10^8} \approx 1.08 \times 10^{-22} \mathrm{kg \cdot m/s} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Photon Wavelength
A photon is a particle of light, and it exhibits both wave-like and particle-like properties. When we talk about the 'photon wavelength,' we're referring to the distance between two successive peaks of the photon's wave. In our exercise, we have a photon with an initial wavelength of 6.13 pm, which is an incredibly short length, showing just how tiny these particles are. The wavelength of a photon is crucial because it determines the energy of the photon. This relationship is given by the equation \[E = \frac{hc}{\text{wavelength}}\]. Understanding photon wavelength helps us in studying various phenomena like Compton Scattering, where a photon changes direction and energy while interacting with an electron. This change involves conservation laws, such as the conservation of momentum and energy.
Scattering Angle
The scattering angle, denoted as \(\theta\), in Compton Scattering refers to the angle by which the photon deviates from its original path after interacting with an electron. This angle is pivotal in determining the change in the wavelength of the scattered photon, which is described by the Compton wavelength shift formula: \[\Delta \lambda = \frac{h}{mc}(1 - \cos\theta)\]. In our exercise, since the problem states that the electron moves in the original direction of the photon, the scattering angle \(\theta\) is 0掳. When \(\theta = 0掳\), \(\cos(0) = 1\), leading to \(\Delta \lambda = 0\). Therefore, there is no change in wavelength. This specific case simplifies the problem, as no wavelength shift means we can easily apply the conservation laws to find the momentum of the electron.
Conservation of Momentum
Conservation of momentum is a fundamental principle in physics, stating that the total momentum of an isolated system remains constant if no external forces act upon it. In the context of Compton Scattering, both the photon and electron before and after the interaction must obey this law. Initially, the photon has momentum and the electron is at rest. After scattering, the electron moves and the photon might change direction and hence its momentum. Since the wavelength doesn鈥檛 change in our specific case (because \(\theta = 0掳\)), the photon's momentum before and after the interaction is the same. Using\(\text{p} = \frac{E}{c}\), we can calculate the photon's momentum and, knowing that the electron must now carry this momentum, we find that the electron's momentum is \(\approx 1.08 \times \10^{-22} \text{kg \cdot m/s}\).
Energy of Photon
The energy of a photon is intimately tied to its wavelength. Using the formula \(\text{E} = \frac{hc}{\lambda}\), where \(h\) is Planck's constant and \(c\) is the speed of light, we can calculate the energy. For a photon with a wavelength of 6.13 pm, the energy comes out to be approximately \(3.24 \times \10^{-14} \text{J}\). This energy is a small value, reflecting the incredibly tiny scales at which photon energy operates. In Compton Scattering, this energy is transferred to the electron upon collision. Thus, understanding the energy of the photon before interaction helps us determine the subsequent kinetic energy and momentum of the electron.
