Problem 68 - Chemistry: An Atoms-Focused Approach

Question

Calculate the energy released and the wavelength of the two photons emitted in the annihilation of an electron and a positron.

Step-by-Step Solution

Answer

Answer: The energy released during the annihilation is 1.64x10^{-13} Joules, and the wavelength of each photon is 8.07x10^{-21} meters.

Step 1: Calculate the energy released.

The energy (E) is equal to the sum of the energies of the electron and the positron. Using the mass-energy equivalence principle (E = mc^2): E = (mass of electron + mass of positron) × c^2 E = (9.11x10^{-31} kg + 9.11x10^{-31} kg) × (3x10^8 m/s)^2 E = 2 × 9.11x10^{-31} kg × (9x10^{16} m^2/s^2) E = 1.64x10^{-13} J The total energy released is 1.64x10^{-13} Joules.

Step 2: Calculate the energy of each photon.

Since two photons are emitted, the energy of each photon (E_photon) is half of the total energy released: E_photon = E / 2 E_photon = 1.64x10^{-13} J / 2 E_photon = 8.2x10^{-14} J The energy of each photon is 8.2x10^{-14} Joules.

Step 3: Calculate the wavelength of each photon.

Using the energy-momentum relationship for photons (E = h/λ), we can find the wavelength (λ) of each photon: λ = h / E_photon λ = 6.626x10^{-34} Js / 8.2x10^{-14} J λ = 8.07x10^{-21} m The wavelength of each photon is 8.07x10^{-21} meters.

Key Concepts

Mass-Energy Equivalence

The principle of mass-energy equivalence, famously articulated by Einstein's equation \( E = mc^2 \), plays a crucial role in understanding phenomena like electron-positron annihilation. This formula reveals how the mass of an object can be converted into energy and vice versa. It states that the energy \( E \) of a system is equal to the mass \( m \) multiplied by the speed of light \( c \) squared.
In the scenario of electron-positron annihilation, both particles, having mass, are converted entirely into energy upon collision. Each of them has a rest mass of approximately \( 9.11 \times 10^{-31} \, \text{kg} \). Using the mass-energy equivalence formula, the total energy released when these particles annihilate is calculated. This energy release is why high-energy photons, or gamma rays, are produced. Understanding this conversion is essential in fields such as nuclear physics, where mass-energy equivalence explains the massive energy outputs witnessed in nuclear reactions.

Photon Energy

Photon energy refers to the amount of energy carried by a photon, which is the basic unit of light or electromagnetic radiation. In the context of electron-positron annihilation, the energy calculated from mass equivalence is shared equally by the two photons.
The total energy released from the annihilation is \( 1.64 \times 10^{-13} \, \text{Joules} \), which is then divided between two photons. Therefore, each photon carries an energy of \( 8.2 \times 10^{-14} \, \text{Joules} \). Understanding photon energy is pivotal in areas such as quantum mechanics and quantum electrodynamics, where the interactions of photons and electrons underpin many of the concepts behind light and electromagnetic phenomena.

• Photon energy is directly related to its frequency by the formula \( E = h u \), where \( h \) is Planck's constant.
• Higher energy photons correspond with higher frequencies, showcasing an inverse relationship with their wavelength.

Wavelength Calculation

When determining the wavelength of a photon, we use its energy and the relationship between energy and wavelength. The equation \( E = \frac{h}{\lambda} \) links the energy \( E \) of a photon to its wavelength \( \lambda \), where \( h \) is Planck's constant.
This formula can be rearranged to find the wavelength: \( \lambda = \frac{h}{E_{\text{photon}}} \). For the 8.2x10^{-14} Joules of energy carried by each photon from annihilation, the wavelength calculates to \( \lambda = 8.07 \times 10^{-21} \, \text{meters} \).

• Wavelength and energy have an inverse relationship: more energy means a shorter wavelength.
• The calculation of photon wavelength is fundamental in spectroscopy, allowing scientists to deduce the structure and behavior of atomic and subatomic particles by studying light.

Wavelengths in this range typically correspond to gamma rays, illustrating the high-energy nature of the photons resulting from particle annihilation.