Calculating Electrostatic Repulsion in Nuclear Fission

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SUMMARY

The discussion focuses on calculating the electrostatic repulsion force between two spheres formed during nuclear fission, each containing 49 protons and a radius of 6.38x10^-15 m. The relevant equation used is F = k (q1q2/d^2), where k is Coulomb's constant (approximately 9E9 N m²/C²). Participants emphasize the importance of correctly identifying the distance (d) between the centers of the spheres, which is not simply the radius multiplied by two. A common mistake noted is the incorrect calculation of the distance, leading to erroneous force values.

PREREQUISITES
  • Understanding of Coulomb's Law and electrostatic forces
  • Familiarity with nuclear fission processes
  • Basic knowledge of atomic structure, specifically protons and their charges
  • Ability to manipulate scientific notation and units in physics calculations
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  • Study the concept of nuclear fission and its implications in nuclear physics
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snash1057
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Homework Statement


At the point of fission, a nucleus with 98 protons is divided into two smaller spheres, each with 49 protons and a radius of 6.38x10^-15 m. What is the repulsive force pushing these two spheres apart

Hint: d ≠ the radius


Homework Equations


F= k (q1q2/d^2)


The Attempt at a Solution


if the radius isn't d then i don't know what is. i have tried multiplying the radius by 2 because there are now 2 spheres and setting it up like

F= 9E9 (49 x 1.6E-19/4.76E-29^2) but that is incorrect
 
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snash1057 said:
i have tried multiplying the radius by 2 because there are now 2 spheres
Makes sense to me.
and setting it up like

F= 9E9 (49 x 1.6E-19/4.76E-29^2) but that is incorrect
Redo this more carefully.
 

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