Question

In one type of mutual annihilation of an electron and a positron, three \(\gamma\)-ray photons are produced. If each photon has an energy of \(0.3407 \mathrm{MeV}\), what is the mass of the positron in amu? (1 \(\mathrm{amu}=931.5 \mathrm{MeV}\) ) (a) \(7.986 \times 10^{-4}\) (b) \(5.486 \times 10^{-4}\) (c) \(16.86 \times 10^{-4}\) (d) \(2.243 \times 10^{-4}\)

Short answer

None of the available options exactly match the calculated result.

Step-by-step solution

Understand the Problem

In this problem, we know the energy of 3 photons produced by the annihilation of an electron and a positron. We need to find the mass of the positron in atomic mass units (amu).

Calculate Total Energy of Photons

Each photon has an energy of 0.3407 MeV. Since there are three photons produced, the total energy is: \[ 3 \times 0.3407 \, \text{MeV} = 1.0221 \, \text{MeV} \]

Use Mass-Energy Equivalence

The equation \( E = mc^2 \) relates energy \( E \) and mass \( m \). Here, we equate the total energy of the photons to the initial mass-energy of the electron-positron pair, specifically the positron's rest mass: \[ m = \frac{E}{c^2} = \frac{1.0221 \, \text{MeV}}{c^2} \]

Convert MeV to amu

Since we need the mass in amu and 1 amu corresponds to 931.5 MeV, convert the result: \[ m_{\text{amu}} = \frac{1.0221 \, \text{MeV}}{931.5 \, \text{MeV/amu}} \approx 1.097 \times 10^{-3} \, \text{amu} \]

Find Closest Answer

Compare the calculated mass of the positron in amu with the options given in the problem: - \(7.986 \times 10^{-4}\)- \(5.486 \times 10^{-4}\)- \(16.86 \times 10^{-4}\)- \(2.243 \times 10^{-4}\)None of the options exactly match our calculation.

Key concepts

Electron-Positron Annihilation

When a particle such as an electron meets its antimatter counterpart, the positron, they can completely annihilate each other. This process, known as annihilation, results in the conversion of their entire rest mass into energy, commonly in the form of photons, specifically gamma (\( \gamma \)-ray) photons. Interestingly, while matter and antimatter do not spontaneously interact in common environments on Earth, this type of interaction is fundamental in particle physics.

• The electron has a negative charge, while the positron has a positive charge, making them perfect mutual annihilators.
• In electron-positron annihilation, energy conservation dictates that the rest mass energy of the particles is converted entirely into energy.
• The number and energy of photons produced can vary, but the total energy of the photons is equal to the total rest mass energy of the electron and positron.

In the example provided, three gamma-ray photons are created, each with an energy of 0.3407 MeV. This demonstrates the principle that even light particles like electrons and positrons can yield significant energy quantities upon annihilation.

Photon Energy Calculation

Photon energy calculations are critical in understanding processes like particle annihilation. The energy of photons is measured in mega-electronvolts (MeV) in the field of particle physics. Here, the scenario involved calculating the total energy of photons produced by an electron-positron annihilation.

• Each photon, in this case, was known to have an energy of 0.3407 MeV.
• To find the total photon energy, you multiply the energy of a single photon by the number of photons:
• \[ 3 \times 0.3407 \, \text{MeV} = 1.0221 \, \text{MeV} \]

This total energy represents the entire energy released in the form of gamma photons from the annihilation. Such calculations are essential for ensuring energy conservation and understanding the energy balance in particle interactions.

Atomic Mass Unit (amu) Conversion

An important step in problems involving mass-energy equivalence is converting between units suitable for the calculations and analysis involved. The atomic mass unit (amu) is commonly used in chemistry and particle physics when dealing with small masses. It is closely related to energy units via Einstein's famous equation, \( E=mc^2 \).
The relationship between these units is expressed as:

• 1 atomic mass unit (amu) equivalently corresponds to 931.5 MeV.
• To find mass from energy in terms of amu, use the conversion formula:
• \[ m_{\text{amu}} = \frac{E}{931.5 \, \text{MeV/amu}} \]

In the exercise provided, the calculated total gamma-ray energy was 1.0221 MeV. By applying the conversion formula: \[ m_{\text{amu}} = \frac{1.0221 \, \text{MeV}}{931.5 \, \text{MeV/amu}} \approx 1.097 \times 10^{-3} \, \text{amu} \] This conversion is crucial to express the results in a more familiar and useful format for comparison and further computations.