Alpha particle approaching gold nucleus

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SUMMARY

The discussion centers on calculating the minimum kinetic energy required for an alpha particle to approach a gold nucleus to a distance of 50 femtometers (fm). An alpha particle consists of two protons and two neutrons, carrying a charge of +2e, while the gold nucleus contains 79 protons with a charge of +79e. The correct formula to use for this calculation is k = 2Ze² / (4πE₀r), where Z is the charge of the nucleus, E₀ is the permittivity of free space, and r is the distance from the nucleus. The rest mass of the gold nucleus is approximately fifty times greater than that of the alpha particle.

PREREQUISITES
  • Understanding of nuclear physics concepts, specifically alpha particles and nuclear charge.
  • Familiarity with electrostatic potential energy equations.
  • Knowledge of fundamental constants such as the electron charge (e) and the permittivity of free space (E₀).
  • Basic grasp of units in nuclear physics, particularly femtometers (fm).
NEXT STEPS
  • Study the derivation of electrostatic potential energy in nuclear interactions.
  • Explore the properties and behavior of alpha particles in nuclear reactions.
  • Learn about the significance of permittivity of free space (E₀) in electrostatics.
  • Investigate the implications of mass differences in nuclear physics, particularly in alpha decay.
USEFUL FOR

This discussion is beneficial for physics students, nuclear physicists, and researchers interested in nuclear interactions and energy calculations involving alpha particles and heavy nuclei.

gilson.fisica
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An alpha particle is thrown toward a gold nucleus. An alpha particle has two neutrons and two protons and a load module 2e, where "e" is the electron charge, while the gold nucleus has 79 protons and a load module 79e. What is the minimum kinetic energy that the alpha particle must have in order to approach to a distance of 50fm the center of the gold nucleus? Assume that the gold core, which rest mass is fifty times greater than the rest mass of the alpha particle.

It's okay to use the equation: k = 2Ze^2/1.pi.Eo.r ?? E = epsilon
Where this is not suitability, know not solve. Help please

Thank...
 
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Apart from the formatting error, the approach looks good, but I think the prefactor is wrong (should be 4 pi in the denominator).

Edit: I changed the topic to give a better description of the problem.
 
Last edited:
Really there was a typing error; k = 2Ze^2/4.pi.Eo.r... Thank you! :)
 

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