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Chapter 39: Problem 59

An electron-positron pair, traveling toward each other with a speed of \(0.99 c\) with respect to their center of mass, collide and annihilate according to \(e^{-}+e^{+} \rightarrow \gamma+\gamma\). Assuming the observer is at rest with respect to the center of mass of the electron-positron pair, what is the wavelength of the photons?

Short Answer

Expert verified
Answer: The wavelength of the photons produced in the electron-positron annihilation is approximately \(1.24 \times 10^{-12} m\).

Step by step solution

01

Write down the energy-momentum relations for the electron-positron pair and the photons

Write down the relativistic energy-momentum relation for the electron, positron, and the photons, using their respective masses \(m_e\), speeds \(v_e\), and energies \(E_e\), \(E_{e^+}\), and \(E_\gamma\): $$E_e = \frac{m_e c^2}{\sqrt{1 - v_e^2/c^2}}$$ $$E_{e^+} = E_e$$ $$E_\gamma = \frac{hc}{\lambda}$$
02

Apply conservation of energy and momentum

Apply the conservation of energy and momentum, stating that the total energy before the collision is equal to the total energy after the collision, and the total momentum before the collision is equal to the total momentum after the collision: $$E_e + E_{e^+} = 2E_\gamma$$ $$E_e v_e - E_{e^+} v_e = 0$$
03

Solve for the wavelength of the photons

Solve for the wavelength \(\lambda\) of the photons produced in the annihilation, substituting the expressions for the energies and speeds in the conservation equations: $$\frac{m_e c^2}{\sqrt{1 - v_e^2/c^2}} + \frac{m_e c^2}{\sqrt{1 - v_e^2/c^2}} = \frac{2hc}{\lambda}$$ $$\frac{2m_e c^2}{\sqrt{1 - v_e^2/c^2}} = \frac{2hc}{\lambda}$$ Now solve for \(\lambda\): $$\lambda = \frac{hc\sqrt{1 - v_e^2/c^2}}{m_e c^2}$$
04

Substitute the given values and calculate the wavelength

Substitute the known values \(v_e = 0.99c\), \(h = 6.626 \times 10^{-34} Js\), \(c = 3 \times 10^8 m/s\), and \(m_e = 9.109 \times 10^{-31} kg\) into the expression for \(\lambda\): $$\lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)\sqrt{1 - (0.99)^2}}{9.109 \times 10^{-31}(3 \times 10^8)}$$ Calculate the wavelength \(\lambda\) to find: $$\lambda \approx 1.24 \times 10^{-12} m$$ The wavelength of the photons produced in the electron-positron annihilation is approximately \(1.24 \times 10^{-12} m\).

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Most popular questions from this chapter

The fundamental observation underlying the Big Bang theory of cosmology is Edwin Hubble's 1929 discovery that the arrangement of galaxies throughout space is expanding. Like the photons of the cosmic microwave background, the light from distant galaxies is stretched to longer wavelengths by the expansion of the universe. This is not a Doppler shift: Except for their local motions around each other, the galaxies are essentially at rest in space; it is the space itself that expands. The ratio of the wavelength of light \(\lambda_{\text {rec }}\) Earth receives from a galaxy to its wavelength \(\lambda_{\text {emit }}\) at emission is equal to the ratio of the scale factor (e.g., radius of curvature) \(a\) of the universe at reception to its value at emission. The redshift \(z\) of the light-which is what Hubble could measure - is defined by \(1+z=\lambda_{\text {rec }} / \lambda_{\text {emit }}=a_{\text {rec }} / a_{\text {emit }}\). a) Hubble's Law states that the redshift \(z\) of light from a galaxy is proportional to the galaxy's distance from us (for reasonably nearby galaxies): \(z \cong c^{-1} H \Delta s\), where \(c\) is the vacuum speed of light, \(H\) is the Hubble constant, and \(\Delta s\) is the distance of the galaxy. Derive this law from the first relationships stated in the problem, and determine the Hubble constant in terms of the scale-factor function \(a(t)\). b) If the present Hubble constant has the value \(H_{0}=72(\mathrm{~km} / \mathrm{s}) / \mathrm{Mpc},\) how far away is a galaxy, the light from which has redshift \(z=0.10\) ? (The megaparsec \((\mathrm{Mpc})\) is a unit of length equal to \(3.26 \cdot 10^{6}\) light-years. For comparison, the Great Nebula in Andromeda is approximately 0.60 Mpc from us.)

Which of the following formed latest in the universe? a) quarks b) protons and neutrons c) hydrogen atoms d) helium nuclei e) gluons

Within three years after it begins operation, the proton beam at the Large Hadron Collider at CERN is expected to reach a luminosity of \(10^{34} \mathrm{~cm}^{-2} \mathrm{~s}^{-1}\) (this means that in a \(1-\mathrm{cm}^{2}\) area, \(10^{34}\) protons encounter each other every second). The cross section for collisions, which could lead to direct evidence of the Higgs boson, is approximately \(1 \mathrm{pb}\) (picobarn). [These numbers were obtained from "Introduction to LHC physics," by G. Polesello, Journal of Physics: Conference Series \(53(2006), 107-116 .]\) If the accelerator runs without interruption, approximately how many of these Higgs events can one expect in one year at the LHC?

A proton and neutron interact via the strong nuclear force. Their interaction is mediated by a particle called a meson, much like the interaction between charged particles is mediated by photons-the particles of the electromagnetic field. a) Perform a rough estimate of the mass of the meson from the uncertainty principle and the known dimensions of a nucleus \(\left(\sim 10^{-15} \mathrm{~m}\right)\). Assume the meson travels at relativistic speed. b) Use a line of reasoning similar to the one in part (a) to prove that the theoretically expected rest mass of the photon is zero. \(.\)

A free neutron decays into a proton and an electron (and an anti-neutrino). A free proton has never been observed to decay into anything. Why then do we consider the neutron to be as "fundamental" (at the nuclear level) a particle as the proton? Why do we not consider a neutron to be a proton-electron composite?
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