The Force Between Alpha Particles at 5.6E-15m

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SUMMARY

The discussion centers on calculating the force between two alpha particles at a distance of 5.6E-15m, specifically within the context of the unstable nucleus of beryllium-8 (8Be). The user applied Coulomb's law, F = k * q1 * q2 / r^2, using the charge of an alpha particle as 1.60E-19 C. However, the calculated force of 41.1N is incorrect due to the misunderstanding of the effective charge of the alpha particles, which is actually +2e (where e is the elementary charge). The correct calculation should use q1 and q2 as 2 * 1.60E-19 C.

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  • Coulomb's Law for electrostatic force calculation
  • Understanding of alpha particle composition (2 protons and 2 neutrons)
  • Knowledge of elementary charge (e = 1.60E-19 C)
  • Basic nuclear physics concepts related to particle stability
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  • Review Coulomb's Law and its applications in particle physics
  • Study the properties of alpha particles and their interactions
  • Learn about nuclear stability and decay processes in isotopes
  • Explore advanced topics in quantum mechanics related to nuclear forces
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Physics students, nuclear physicists, and anyone interested in understanding forces between subatomic particles and nuclear stability.

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The nucleus of 8Be, which consists of 4 protons and 4 neutrons is very unstable and separates into two alpha particles (2 protons and 2 neutrons) what is the force between the two alpha particles if the distance is 5.60*10^-15m

I thought you would just use the formula F=ke*q1*q2/r^2

so I did that and I repeatedly have the wrong answer:

F= (8.99*10^9)(1.60*10^-19)(1.60*10^-19)/(5.60*10^-15)^2= 41.1N

Please tell what am I doing wrong?
 
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