A positron and an electron can annihilate each other on colliding, producing energy as photons: $$ { }_{-1}^{0} \mathrm{e}+{ }_{+1}^{0} \mathrm{e} \longrightarrow 2{ }^{0}{ }_{0}^{0} \gamma $$ Assuming that both \(\gamma\) rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

Short Answer

Expert verified
The wavelength of each gamma ray produced in the annihilation process of a positron and an electron is approximately \(1.21 \times 10^{-12}\) meters.

Step by step solution

01

Determine the total energy released after annihilation

We are going to use Einstein's mass-energy equivalence formula, which states that the energy of an object is given by the product of its mass and the speed of light squared: E = mc^2 where E is the energy, m is the mass, and c is the speed of light. Both electron and positron have the same mass (m_e = 9.11 x 10^-31 kg), and the total energy (E_total) released in the process is the sum of their energies: E_total = E(positron) + E(electron) = 2 * m_e * c^2
02

Calculate the energy of the produced gamma rays

Given that both gamma rays have the same energy, the energy per gamma ray (E_gamma) can be calculated by dividing the total energy by 2: E_gamma = E_total / 2
03

Use the energy-wavelength relationship to find the wavelength

Now we can use the energy-wavelength relationship, which is given by the Planck's equation: E = h * c / λ where E is the energy, h is the Planck's constant (6.63 x 10^-34 Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength. We can rearrange the formula to solve for λ: λ = h * c / E_gamma Now plug in the values for E_gamma, h, and c to calculate the wavelength: λ = (6.63 x 10^-34 Js) * (3 x 10^8 m/s) / E_gamma
04

Calculate the final wavelength

First, we need to find E_gamma. We have the E_total = 2 * m_e * c^2 and E_gamma = E_total / 2: E_gamma = (2 * (9.11 x 10^-31 kg) * (3 x 10^8 m/s)^2) / 2 Now we plug in E_gamma value into the formula for λ: λ = (6.63 x 10^-34 Js) * (3 x 10^8 m/s) / E_gamma After calculating, we get the value of λ: λ ≈ 1.21 x 10^-12 m So, the wavelength of each gamma ray produced in the annihilation process is approximately 1.21 x 10^-12 meters.

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Most popular questions from this chapter

Which of the following statement(s) is(are) true? a. A radioactive nuclide that decays from \(1.00 \times 10^{10}\) atoms to \(2.5 \times 10^{9}\) atoms in 10 minutes has a half-life of 5.0 minutes. b. Nuclides with large \(Z\) values are observed to be \(\alpha\) -particle producers. c. As \(Z\) increases, nuclides need a greater proton-to-neutron ratio for stability. d. Those "light" nuclides that have twice as many neutrons as protons are expected to be stable.

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