Question

When a positron is annihilated by combination with an electron, two photons of equal energy result. What is the wavelength of these photons? Are they gamma ray photons?

Short answer

The wavelength of the photons produced when a positron is annihilated by combining with an electron is approximately \(2.42\times10^{-12} m\). Since this wavelength is less than \(10^{-11} m\), these photons are considered gamma-ray photons.

Step-by-step solution

Find the energy released during the annihilation

When a positron and electron annihilate each other, their total mass is converted into energy, which is released as two photons of equal energy. We can use the energy-mass equivalence formula to find the energy released: \[E = mc^2\] where E is the energy, m is the combined mass of the positron and electron, and c is the speed of light (approximately \(3\times10^{8} m/s\)). The mass of an electron (and positron, since they have the same mass) is approximately \(9.11\times10^{-31}kg\). Therefore, the combined mass m is: \[m = 2\times(9.11\times10^{-31}kg) = 1.822\times10^{-30}kg\] Now, we can find the energy released (E) by the annihilation: \[E = (1.822\times10^{-30}kg)(3\times10^8 m/s)^2 = 1.64\times10^{-13}J\]

Determine the energy of a single photon

Since the annihilation process creates two photons with equal energy, we can calculate the energy of a single photon by dividing the total energy released (E) by 2: \[E_{photon} = \frac{1.64\times10^{-13}J}{2} = 8.2\times10^{-14}J\]

Calculate the frequency of the photon

We can use the energy of a photon formula to find the frequency (f) of the photon: \[E_{photon} = hf\] where h is the Planck's constant (\(6.63\times10^{-34} Js\)). Rearranging the formula, we get: \[f = \frac{E_{photon}}{h}\] Now we can plug in the values and find the frequency: \[f = \frac{8.2\times10^{-14}J}{6.63\times10^{-34}Js} \approx 1.24\times10^{20} Hz\]

Calculate the wavelength of the photon

Now we can use the speed of light, frequency, and the formula \(c = f\lambda\) to find the wavelength (λ) of the photon: \[\lambda = \frac{c}{f}\] Plugging in the values, we get: \[\lambda = \frac{3\times10^8 m/s}{1.24\times10^{20} Hz} \approx 2.42\times10^{-12} m\]

Determine if the photon is a gamma-ray photon

Gamma rays have wavelengths typically less than \(10^{-11} m\). Since the calculated wavelength of our photon is approximately \(2.42\times10^{-12} m\), which is less than \(10^{-11} m\), we can conclude that these photons are gamma-ray photons.

Key concepts

Energy-Mass Equivalence

The energy-mass equivalence principle is a fundamental concept introduced by Albert Einstein as part of his theory of relativity. It's often summed up by the famous equation, E = mc^2, where E stands for energy, m is mass, and c represents the speed of light in a vacuum. In essence, this equation tells us that mass can be converted into energy and vice versa. The idea of mass-energy equivalence is perfectly illustrated in the process of electron-positron annihilation.

When an electron and a positron collide, they annihilate each other, converting their entire mass into energy. This energy manifests as photons, specifically gamma-ray photons in most cases. The energy released can be calculated by simply doubling the mass of one of the particles, since an electron and a positron have equal mass, and then applying the E = mc^2 equation. This process is a direct demonstration of how matter can produce a considerable amount of energy, even when starting with something as tiny as electron and positron particles.

Photon Energy

Photon energy is related to its electromagnetic properties and is fundamentally important when discussing light and other forms of electromagnetic radiation. The energy of a single photon can be determined using the formula E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon. By increasing the frequency, we increase the energy, while longer wavelengths correspond to photons with less energy.

For example, in the case of electron-positron annihilation, the energy released from this transformation creates two photons of equal energy. This energy can be found by applying the energy-mass equivalence principle first to determine the total energy released, and then dividing by two to find the individual photon energy. The frequency and, correspondingly, the wavelength of these photons can then be calculated, which in turn can identify the type of radiation we're dealing with, whether it's radio waves, visible light, or in our case, gamma rays.

Gamma Rays

Gamma rays represent the highest-energy end of the electromagnetic spectrum and are characterized by their extremely short wavelengths, typically less than 10^-11 meters. These rays have enough energy to ionize atoms and molecules, which can damage biological tissue and are therefore used with great care in medical treatments and sterilization processes.

When we talk about gamma rays in the context of electron-positron annihilation, the photons produced during annihilation have a very short wavelength, indicating they are indeed gamma rays. This also suggests that the process releases a significant amount of energy, given the high frequency and high energy of these photons. The presence of gamma rays is a key indication of the high-energy processes occurring at a subatomic level, such as in nuclear reactions or in particle-antiparticle annihilation events, further underscoring the importance of understanding such high-energy phenomena in physics.