Problem 25 - University Physics with Modern Physics

Question

An electron and a positron are moving toward each other and each has speed 0.500\(c\) in the lab frame. (a) What is the kinetic energy of each particle? (b) The e' and e^ - meet head-on and andhilate. What is the energy of each photon that is produced? (c) What is the wavelength of each photon? How does the wavelength compare to the photon wavelength when the initial kinetic energy of the \(\mathrm{e}^{+}\) and \(\mathrm{e}^{-}\) is negligibly small (see Example 38.6\() ?\)

Step-by-Step Solution

Answer

Each photon energy is 0.5898 MeV, and the wavelength is approximately 2.11 pm, shorter than if the initial kinetic energies were negligible.

Step 1: Calculate Rest Energy

The rest energy of an electron (or positron) is given by the equation \( E_0 = mc^2 \), where \( m \) is the rest mass of the electron (approximately \( 9.11 \times 10^{-31} \ ext{kg} \)) and \( c \) is the speed of light (approximately \( 3 \times 10^8 \ ext{m/s} \)). The rest energy thus calculates to approximately \( 0.511 \ ext{MeV} \).

Step 2: Calculate Kinetic Energy

The kinetic energy for each particle is given by \( KE = (rac{1}{\_\sqrt{1-(v/c)^2}} - 1)mc^2 \). With \( v = 0.5c \), the Lorentz factor (\( \gamma \)) computes to \( 1.154 \). Substitute into the kinetic energy formula to find \( KE \approx 0.154 \times 0.511 \ ext{MeV} = 0.0788 \ ext{MeV} \).

Step 3: Calculate Total Energy Available

Upon annihilation, the total energy is the sum of the rest energy and kinetic energy of both particles. Each particle contributes \( E = E_0 + KE = 0.511 \ ext{MeV} + 0.0788 \ ext{MeV} = 0.5898 \ ext{MeV} \). Since there are two particles, the total energy is \( 2 \times 0.5898 \ ext{MeV} = 1.1796 \ ext{MeV} \).

Step 4: Calculate Energy of Each Photon

In a head-on annihilation, the total energy is distributed equally among the photons produced. Therefore, each photon has energy \( E_{\gamma} = \frac{1.1796}{2} = 0.5898 \ ext{MeV} \).

Step 5: Calculate Wavelength of Each Photon

The wavelength \( \lambda \) of a photon is given by \( \lambda = \frac{hc}{E} \), where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \ ext{J s} \)) and \( E \) is the photon energy converted to joules. Substitute \( E = 0.5898 \ ext{MeV} \) as joules and calculate \( \lambda \approx 2.11 \times 10^{-12} \ ext{m} \).

Step 6: Compare Wavelength with Negligible Kinetic Energy Case

When the initial kinetic energies are negligible, each photon takes the rest energy of an electron or positron (\( 0.511 \ ext{MeV} \)), leading to a larger wavelength \( \lambda_{0} = \frac{hc}{0.511 \ ext{MeV}} \). Compare \( \lambda_{0} \) and calculate it to be larger than the current wavelength, reflecting less energy per photon in this scenario.

Key Concepts

Kinetic Energy

Kinetic energy (KE) is the energy that an object possesses due to its motion. For particles like electrons and positrons moving at high speeds, such as 0.500 times the speed of light, KE becomes significantly interesting. In the realm of special relativity, the formula to compute the kinetic energy of a moving particle involves the Lorentz factor \( \gamma \), which accounts for relativistic effects.

The relativistic kinetic energy is calculated using:

• \( KE = \left( \frac{1}{\sqrt{1-(v/c)^2}} - 1 \right) mc^2 \)

Here, \( m \) is the rest mass of the particle, and \( v \) is its velocity. For the specified speed of 0.500\( c \), the KE of an electron equates to roughly 0.0788 MeV. Understanding this value is crucial as it signifies how much energy is involved when high-speed particles eventually collide or interact.

Photon Energy

Photon energy refers to the energy carried by a photon, a particle of light. During electron-positron annihilation, all the energy, including the particles' rest and kinetic energy, converts into photon energy. This process ensures energy conservation, which is a fundamental principle of physics.

In the given problem, each resulting photon from the annihilation carries an energy of 0.5898 MeV. This calculation stems from dividing the total energy by the two photons produced in the head-on annihilation:

• \( E_{\gamma} = \frac{1.1796 \text{ MeV}}{2} \)

Understanding photon energy is essential as it links directly to other aspects such as frequency and wavelength, influencing photon interactions with matter.

Wavelength Calculation

To understand light and electromagnetic radiation, calculating the wavelength of photons is essential. The wavelength \( \lambda \) can be derived using the equation:

• \( \lambda = \frac{hc}{E} \)

Where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ Js} \)), \( c \) is the speed of light, and \( E \) is the energy of the photon. This formula reveals that as photon energy increases, the wavelength decreases.

In the exercise, after determining the photon energy as 0.5898 MeV, converting it to joules allows the calculation of the wavelength, approximately \( 2.11 \times 10^{-12} \text{ m} \). Comparing this to scenarios with negligible kinetic energy shows how initial particle speeds affect photon wavelengths.

Special Relativity

Special relativity is crucial when dealing with objects moving at speeds close to the speed of light. It corrects Newtonian mechanics using transformations that consider high velocities' effects. One cornerstone of special relativity is the relation between energy, mass, and speed expressed as \( E = mc^2 \).

In our scenario, the kinetic and rest energy of particles are integrated under special relativity principles to compute photon energy accurately. This framework shifts our understanding of how mass and energy interplay at relativistic velocities, enabling precise predictions about particle interactions.

Lorentz Factor

The Lorentz factor \( \gamma \) plays a key role in describing how time, length, and relativistic mass change for an object in motion. It is defined as:

• \( \gamma = \frac{1}{\sqrt{1-(v/c)^2}} \)

This factor adjusts calculations of kinetic energy and momentum when speeds approach the speed of light.

In the exercise, with \( v = 0.5c \), \( \gamma \) is computed to be roughly 1.154. This adjustment reflects how velocities impact other physical quantities. Mastering the concept of the Lorentz factor is significant in relativistic physics as it helps explain time dilation, length contraction, and kinetic energy calculations in high-speed scenarios.