Alpha particle approaching gold nucleus

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An alpha particle, consisting of two protons and two neutrons, approaches a gold nucleus with 79 protons, requiring a minimum kinetic energy to reach within 50 femtometers of the nucleus. The relevant equation for calculating the necessary kinetic energy is k = 2Ze^2/4πEo.r, where Z represents the charge of the nucleus and r is the distance from the center. The discussion highlights a correction in the equation's prefactor from 1 to 4 in the denominator. The rest mass of the gold nucleus is noted to be significantly greater than that of the alpha particle, influencing the dynamics of the interaction. Overall, the focus is on determining the kinetic energy threshold for the alpha particle's approach to the gold nucleus.
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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.
 
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Really there was a typing error; k = 2Ze^2/4.pi.Eo.r... Thank you! :)
 
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