A photon of energy 4.1 MeV is incident on a lead nucleus., causing the creation of an electron-positron pair. They travel perpendicular to the initial direction of the photon.The energy of the electron is_
Correct answer: B. 2.05 MeV
- A. 4.1 MeV
- B. 2.05 MeV
- C. 1.02 MeV
- D. 0.51 MeV
Explanation
We know, that in pair production momentum and energy are conserved. By conserving energy, E total = Ee + Ep 4.1 MeV = Ee + Ep Now, by conserving momentum in the y-direction we get Pe - Pp = 0 Pe = Pp Hence, if momentum is the same the energy should be the same Therefore, 4.1 MeV = 2 Ee Ee= 2.05 MeV.
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About Photons and the Particle Model of Light
The particle model treats light as photons with energy E equals hf and momentum related to wavelength. Questions involve the photoelectric effect, work function, threshold frequency, stopping potential and photon energy, while distinguishing particle behaviour from the wave description of interference and diffraction.
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