The question asks about the minimum energy released during the annihilation of an electron and a positron. This phenomenon is a direct consequence of Einstein's mass-energy equivalence principle, where mass is converted into energy.

Positron Annihilation: Understanding the Basics

Annihilation occurs when a particle and its corresponding antiparticle meet. In this case, we have an electron and its antiparticle, a positron.

• Electron: A fundamental particle with a negative charge and a specific rest mass. Its symbol is \(e^-\).
• Positron: The antiparticle of the electron. It has the same mass as an electron but carries a positive charge. Its symbol is \(e^+\).

When an electron and a positron come into contact, they annihilate, meaning they cease to exist as particles. Their entire mass is converted into energy, primarily in the form of high-energy photons (gamma rays). The minimum energy released corresponds to the conversion of their combined rest mass into energy.

Energy Release through Mass-Energy Equivalence

The energy released from the conversion of mass is governed by Einstein's famous equation:

\[E = mc^2\]

Where:

• \(E\) represents the energy released.
• \(m\) is the total mass that is converted into energy.
• \(c\) is the speed of light in a vacuum, which is approximately \(3.00 \times 10^8\) meters per second (m/s).

Calculating Minimum Energy from Electron-Positron Annihilation

To determine the minimum energy released, we need to consider the total rest mass of the electron-positron pair. The rest mass of a single electron (or positron) is a known fundamental constant.

Rest mass of an electron (\(m_e\)) is approximately: \(9.109 \times 10^{-31}\) kg

Since both the electron and the positron annihilate, the total mass (\(m_{total}\)) that is converted into energy is the sum of their individual rest masses:

\[m_{total} = m_{electron} + m_{positron}\]

As the mass of an electron is equal to the mass of a positron (\(m_e\)), we have:

\[m_{total} = m_e + m_e = 2m_e\]

Now, let's substitute the value of \(m_e\) into this equation:

\[m_{total} = 2 \times (9.109 \times 10^{-31} \text{ kg}) = 1.8218 \times 10^{-30} \text{ kg}\]

Next, we apply Einstein's mass-energy equivalence formula:

\[E = m_{total} c^2\]

\[E = (1.8218 \times 10^{-30} \text{ kg}) \times (3.00 \times 10^8 \text{ m/s})^2\]

\[E = (1.8218 \times 10^{-30}) \times (9.00 \times 10^{16}) \text{ J}\]

\[E = 16.3962 \times 10^{-14} \text{ J}\]

Converting Energy to Mega-electron Volts (MeV)

The options provided are in Mega-electron Volts (MeV), so we need to convert the energy from Joules to MeV. The necessary conversion factors are:

• \(1 \text{ electron-volt (eV)} = 1.602 \times 10^{-19} \text{ J}\)
• \(1 \text{ Mega-electron-volt (MeV)} = 10^6 \text{ eV}\)

First, convert the energy from Joules to electron-volts:

\[E \text{ (eV)} = \frac{16.3962 \times 10^{-14} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}}\]

\[E \text{ (eV)} \approx 10.234 \times 10^5 \text{ eV}\]

Now, convert electron-volts to Mega-electron-volts:

\[E \text{ (MeV)} = \frac{10.234 \times 10^5 \text{ eV}}{10^6 \text{ eV/MeV}}\]

\[E \text{ (MeV)} \approx 1.0234 \text{ MeV}\]

Rounding to two decimal places, the minimum energy released is approximately \(1.02\) MeV.

Conclusion: Minimum Energy Released from Annihilation

The calculation clearly shows that the minimum energy released when an electron and a positron annihilate is approximately \(1.02\) MeV. This energy is typically emitted as two gamma-ray photons, with each photon carrying about \(0.51\) MeV of energy, moving in opposite directions to conserve momentum.

The final calculated value is 1.02 MeV.