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Chapter 11: Problem 33

When both an electron and a positron are at rest, they can annihilate each other according to the reaction \(e^{-}+e^{+} \rightarrow \gamma+\gamma\). In this case, what are the energy, momentum, and frequency of each photon?

Short Answer

Expert verified
In the electron-positron annihilation, both photons will have equal energy, momentum, and frequency. The energy of each photon is \(E_{photon} = m_e c^2\), where \(m_e\) is the electron rest mass and \(c\) is the speed of light. The momentum of each photon is \(p_{photon} = m_e c\). The frequency of each photon is \(\nu = \frac{m_e c^2}{h}\), where \(h\) is the Planck's constant.

Step by step solution

01

Identify the given information

We are given that an electron and a positron, both initially at rest, annihilate each other to produce two photons.
02

Use conservation of energy

According to the conservation of energy, the total energy before and after the annihilation must be equal. The initial energy of the system is the sum of the rest energies of the electron and positron. Since both particles have the same rest mass \(m_e\), their total initial energy is given by \(E_{initial} = 2 m_e c^2\), where \(c\) is the speed of light. The final energy of the system is the sum of the energies of the two photons produced in the annihilation. Since both photons have the same energy, we can say \[E_{photon}=\frac{1}{2}E_{initial}\] \[E_{photon}=\frac{1}{2}(2m_e c^2)\] \[E_{photon}=m_e c^2\]
03

Use conservation of momentum

According to the conservation of momentum, the total momentum before and after the annihilation must be equal. Since both the electron and positron are initially at rest, their total initial momentum is zero. To conserve momentum, the two photons produced must have equal and opposite momenta, meaning their magnitudes are equal.
04

Determine the momentum of the photons

To determine the momentum of the photons, we use the energy-momentum relation for photons: \[E=pc\] where E is the photon energy, p is the photon momentum, and c is the speed of light. We obtained the energy of each photon in step 2 as \(E_{photon} = m_e c^2\). Therefore, the momentum of each photon is given by: \[p_{photon}=\frac{E_{photon}}{c}\] \[p_{photon} = \frac{m_e c^2}{c}\] \[p_{photon} = m_e c\]
05

Determine the frequency of the photons

Finally, we use Planck's formula to relate the energy of the photons to their frequency: \[E=h\nu\] where \(E\) is the photon energy, \(h\) is the Planck's constant, and \(\nu\) is the frequency of the photon. We have the energy of each photon from step 2 as \(E_{photon} = m_e c^2\). Therefore, the frequency of each photon is given by: \[\nu = \frac{E_{photon}}{h}\] \[\nu = \frac{m_e c^2}{h}\] So the energy, momentum, and frequency of each photon produced in the electron-positron annihilation are \(m_e c^2\), \(m_e c\), and \(\frac{m_e c^2}{h}\), respectively.

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