Problem 74 - Chemistry

Question

A positron and an electron can annihilate each other on colliding, producing energy as photons: $$ -_{-1}^{0} e+_{+1}^{0} e \longrightarrow 2_{0}^{0} \gamma $$ Assuming that both g rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

Step-by-Step Solution

Answer

The wavelength of the electromagnetic radiation produced during the annihilation of a positron and an electron can be calculated with the following formula: \(\lambda = \frac{hc}{E_{photon}}\), where \(h = 6.626 \times 10^{-34} Js\) is Planck's constant, \(c = 3.0 \times 10^8 m/s\) is the speed of light, and \(E_{photon}\) is the energy of a single photon. Substituting the given values, we get: \(\lambda = \frac{(6.626 \times 10^{-34} Js)(3.0 \times 10^8 m/s)}{E_{photon}}\).

Step 1: Calculate the total energy produced

Using Einstein's formula, \( E = mc^2 \), for both the electron and the positron: \( E_{electron} = m_{electron}c^2 \) \( E_{positron} = m_{positron}c^2 \) Since the electron and positron have the same mass, the total energy produced will be: \( E_{total} = E_{electron} + E_{positron} = 2 m_e c^2 \) Where, \(m_e = 9.109 \times 10^{-31} kg \) is the mass of an electron/positron and \(c = 3.0 \times 10^8 m/s\) is the speed of light.

Step 2: Calculate the energy of a single photon

Since two photons of equal energy are produced during the annihilation, the energy of a single photon is: \( E_{photon} = \frac{E_{total}}{2} \)

Step 3: Calculate the wavelength (\(\lambda\)) of the electromagnetic radiation

Now we will use the energy-wavelength equation: \( E_{photon} = h\frac{c}{\lambda} \) Rearranging for the wavelength: \( \lambda = \frac{hc}{E_{photon}} \) Where, \(h = 6.626 \times 10^{-34} Js\) is the Planck's constant Now substituting the given values for the masses, speed of light, and Planck's constant, and the energy of a single photon that we calculated in step 2: \( \lambda = \frac{(6.626 \times 10^{-34} Js)(3.0 \times 10^8 m/s)}{E_{photon}} \)

Step 4: Calculate the result

Suppose \(m_e = 9.109 \times 10^{-31} kg\), \(c = 3.0 \times 10^8 m/s\), and \(h = 6.626 \times 10^{-34} Js\), then: \(E_{total}=2m_ec^2=2 \times (9.109 \times 10^{-31} kg) \times (3.0 \times 10^8 m/s)^2\) \(E_{photon}=\frac{E_{total}}{2}\) And finally, \( \lambda = \frac{(6.626 \times 10^{-34} Js)(3.0 \times 10^8 m/s)}{E_{photon}} \) This equation will give us the wavelength of the electromagnetic radiation produced during the annihilation of positron and electron.

Key Concepts

Photon Energy

Photons are tiny packets of energy that make up light and electromagnetic radiation. When speaking about photon energy, we're discussing how much energy a single photon carries. This energy is directly related to its frequency and inversely related to its wavelength. In the world of particle physics, particularly during events like an electron and positron annihilation, photon energy plays a key role. These particles annihilate each other resulting in energy released in the form of photons.
Each photon carries energy calculated using the formula:

• \( E_{photon} = \frac{E_{total}}{2} \)

As seen in the physics of annihilation, the electron and positron provide their rest energy to form two photons. Therefore, understanding photon energy helps us to decipher various physical phenomena, including the radiation type produced and its properties. To compute photon energy, one typically requires the total energy from such a collision event, which can then be divided evenly among the photons produced.

Electromagnetic Radiation

Electromagnetic radiation is a form of energy that travels through space, encompassing a broad range of wavelengths from radio waves to gamma rays. Each type of electromagnetic radiation is characterized by its wavelength or frequency, and photons are the particle representation of this phenomenon.
When electron and positron particles collide and annihilate, they produce electromagnetic radiation in the form of gamma rays, which are high-energy, short-wavelength radiation.
Properties of electromagnetic radiation include:

• Traveling at the speed of light \( (c = 3.0 \times 10^8 \text{m/s}) \).
• Ability to travel through a vacuum.
• Carrying energy that can be calculated based on the electromagnetic spectrum.

Understanding electromagnetic radiation is crucial in physics as it helps us explain how particle collisions can produce measurable energies and radiation. This understanding also aids in technological advancements and scientific exploration ranging from medical imaging to understanding outer space.

Einstein's Equation

Einstein's equation, also known as the mass-energy equivalence principle, is famously presented as \( E = mc^2 \), where:

• \( E \) stands for energy.
• \( m \) denotes mass.
• \( c \) is the speed of light in a vacuum.

This equation illustrates that mass can be converted into energy, a concept that is fundamental to nuclear reactions and particle physics. During an electron-positron annihilation, this principle explains how the rest mass of these particles is efficiently transformed into energy.
This conversion is why the result of such annihilation includes the production of high-energy photons. Einstein's equation not only describes a cornerstone of modern physics but also underscores the remarkable potential of matter to release energy. Understanding this equation aids in exploring how small amounts of mass in matter-antimatter collisions can resultantly produce huge amounts of energy in the form of photons.

Energy-Wavelength Relation

The relationship between energy and wavelength is central to understanding why different wavelengths of light have different amounts of energy. This concept is embedded within the equation:

• \( E_{photon} = h\frac{c}{\lambda} \)

where:

• \( E_{photon} \) is the energy of a photon.
• \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{Js})\).
• \( \lambda \) is the wavelength of the photon.

The energy-wavelength relation shows that the shorter the wavelength, the higher the energy of the photon. This is because the wavelength is inversely related to energy, making this principle a powerful tool in the analysis of radiation.
During electron and positron annihilation, photons are emitted with specific wavelengths calculated from their energy, using the stated formula. This relationship is vital in fields such as spectroscopy, which relies on measuring energy to determine wavelength and identify substances.