Question

A positron has the same mass as an electron \(\left(9.109 \times 10^{-31} \mathrm{~kg}\right)\) but an opposite charge. When the two particles encounter each other, annihilation occurs and only \(\gamma\) rays are produced. How much energy (in \(\mathrm{ks} / / \mathrm{mol}\) ) is produced?

Short answer

The energy produced is \(9.879 \times 10^{7}\, \text{ks/mol}\).

Step-by-step solution

Understand the Annihilation Process

In the annihilation process, a positron and an electron collide and transform their entire mass into energy in the form of gamma rays. According to Einstein's mass-energy equivalence principle, the energy produced is given by the formula \( E = mc^2 \), where \( m \) is the mass and \( c \) is the speed of light \( (3 \, \times 10^8 \, \text{m/s}) \).

Calculate Energy for One Pair

The total mass of the positron-electron pair is twice the mass of a single electron, which is \( 2 \times 9.109 \times 10^{-31} \, \text{kg} = 1.8218 \times 10^{-30} \,\text{kg} \). Thus, the energy produced from one pair is \( E = (1.8218 \times 10^{-30} \, \text{kg})(3 \times 10^8 \, \text{m/s})^2 = 1.640 \times 10^{-13} \, \text{J} \).

Convert Joules to Kilojoules per Mole

To convert the energy from joules per particle to kilojoules per mole, use Avogadro's number \( 6.022 \times 10^{23} \, \text{mol}^{-1} \). The energy per mole is \( E_{mol} =(1.640 \times 10^{-13} \, \text{J}) \times (6.022 \times 10^{23} \, \text{mol}^{-1}) = 9.879 \times 10^{10} \, \text{J/mol} \). Converting to kilojoules, \( E_{mol} = 9.879 \times 10^{7} \, \text{kJ/mol} \).

Calculate Energy in ks/mol

Since 1 kJ = 10^3 J and 1 ks = 10^3 s, the unit conversion is straightforward without any need for further calculation in terms of seconds. Thus, the final energy output is \( 9.879 \times 10^{7} \, \text{ks/mol} \).

Key concepts

Positron-Electron Annihilation

When a positron meets an electron, a fascinating event takes place: they annihilate each other. Both particles have the same mass, approximately \(9.109 \times 10^{-31} \text{ kg}\), but carry opposite electric charges. The positron has a positive charge, while the electron has a negative one.
During annihilation, the entire mass of the two particles is converted into energy. This transformation typically results in the production of gamma rays, a type of high-energy photon. This process exemplifies the concept of mass-energy equivalence, where mass can be transformed into energy and vice versa.

- **Particle collision:** When a positron encounters an electron, they collide. - **Mass transformation:** Their combined mass changes form, releasing energy. - **Energy form:** The energy is emitted as gamma rays.

Gamma Rays

Gamma rays are a powerful form of electromagnetic radiation, similar but far more energetic than visible light or X-rays. During positron-electron annihilation, the mass of both particles is converted into gamma rays, which is a clear demonstration of mass-energy equivalence in action. These rays are particularly energetic because they arise from the complete conversion of the particle's rest mass into energy.

- **Nature:** Gamma rays are electromagnetic waves with very short wavelengths and extremely high frequencies. - **Energy:** These rays carry a lot of energy, which is why they are used in technologies that require intense energy outputs. - **Annihilation output:** In the context of positron-electron annihilation, all the energy from the conversion is emitted as gamma rays.

One reason gamma rays are important in physics is their ability to penetrate materials and their role in illustrating fundamental physics concepts. Understanding how gamma rays form and their characteristics gives insight into the massive energy potential present in even small amounts of mass.

Einstein's Mass-Energy Formula

Einstein's renowned mass-energy formula, \( E = mc^2 \), is essential for understanding how mass can be converted into energy, as happens in positron-electron annihilation. This formula shows that even a small amount of mass can be transformed into a significant quantity of energy due to the large value of the speed of light squared \((c^2)\).

- **Formula breakdown:** In \( E = mc^2 \), \(E\) represents energy, \(m\) is mass, and \(c\) is the speed of light, approximately \(3 \times 10^8 \text{ m/s}\). - **Implication:** Small masses can yield huge energies because \(c^2\) is so large.

This equation is at the heart of modern physics and provides the framework for many phenomena, from nuclear reactions to cosmological events. In the context of positron-electron annihilation, it allows us to compute the exact energy released when these particles convert their mass into gamma rays. Understanding this formula helps students grasp why processes that involve even tiny particles can result in substantial energy outputs.