Question

In gamma-ray astronomy, the existence of positrons \(\left(\mathrm{e}^{+}\right)\) can be inferred by a characteristic gamma ray that is emitted when a positron and an electron (e \(^{-}\) ) annihilate. For simplicity, assume that the electron and positron are at rest with respect to an Earth observer when they annihilate and that nothing else is in the vicinity. (a) Consider the reactions $\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow \gamma$, where the annihilation of the two particles at rest produces one photon, and \(\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow 2 \gamma,\) where the annihilation produces two photons. Explain why the first reaction does not occur, but the second does. (b) Suppose the reaction \(\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow 2 \gamma\) occurs and one of the photons travels toward Earth. What is the energy of the photon?

Short answer

Answer: The energy of the photon reaching Earth is \(8.19 \times 10^{-14} \ \mathrm{J}\).

Step-by-step solution

Analyze the given reactions

Given the two reactions: 1. \(e^{-} + e^{+} \rightarrow \gamma\) 2. \(e^{-} + e^{+} \rightarrow 2 \gamma\) We need to determine which reaction occurs while preserving the conservation laws of energy and momentum.

Apply conservation laws to the first reaction

For the first reaction, \(e^{-} + e^{+} \rightarrow \gamma\), let's analyze if it obeys the conservation laws of energy and momentum. Before reaction: Total energy: \(E = E_{e^{-}} + E_{e^{+}} = m_ec^2 + m_ec^2\) Momentum: \(p = p_{e^{-}} + p_{e^{+}} = 0 + 0 = 0\) After reaction: Total energy: \(E = E_{\gamma} = h\nu\) Momentum: \(p = p_{\gamma} = \frac{h\nu}{c}\) The conservation laws are not preserved in this reaction, because the initial total momentum is 0, while the final total momentum is non-zero. Hence, the first reaction does not occur.

Apply conservation laws to the second reaction

For the second reaction, \(e^{-} + e^{+} \rightarrow 2 \gamma\), let's analyze if it obeys the conservation laws of energy and momentum. Before reaction: Total energy: \(E = E_{e^{-}} + E_{e^{+}} = m_ec^2 + m_ec^2\) Momentum: \(p = p_{e^{-}} + p_{e^{+}} = 0 + 0 = 0\) After reaction: Total energy: \(E = E_{\gamma_1} + E_{\gamma_2} = h\nu_1 + h\nu_2\) Momentum: \(p = p_{\gamma_1} + p_{\gamma_2} = \frac{h\nu_1}{c} + \frac{h\nu_2}{c}\) Conservation laws can be preserved in this reaction if the two photons are emitted in opposite directions so that their momenta cancel each other out. Therefore, the second reaction does occur.

Calculate the energy of the photon reaching Earth

To calculate the energy of the photon reaching Earth in the second reaction, \(e^{-} + e^{+} \rightarrow 2 \gamma\), we can use the conservation of energy. Since the electron and positron are at rest and have the same mass, they both contribute equal amounts of rest energy: Total initial energy: \(E = E_{e^{-}} + E_{e^{+}} = m_ec^2 + m_ec^2 = 2m_ec^2\) Restricting the analysis to one of the photons produced, half of the total initial energy corresponds to the energy of that photon: Energy of the photon: \(E_{\gamma} = \frac{1}{2} \times 2m_ec^2 = m_ec^2\) Use the mass of electron \(m_e = 9.11 \times 10^{-31} \ \mathrm{kg}\) and the speed of light \(c = 3 \times 10^8 \ \mathrm{m/s}\). \(E_{\gamma} = (9.11 \times 10^{-31} \ \mathrm{kg})(3 \times 10^8 \ \mathrm{m/s})^2 = 8.19 \times 10^{-14} \ \mathrm{J}\) So, the energy of the photon reaching Earth is \(8.19 \times 10^{-14} \ \mathrm{J}\).