What Forces Act Between Alpha Particles in Unstable Nuclei?

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The discussion focuses on the forces acting between alpha particles in the unstable nucleus of beryllium-8, which breaks into two alpha particles. Users are attempting to calculate the force between the particles when they are 5 x 10^-15 m apart using Coulomb's law, but one participant is struggling to arrive at the correct answer of 36.8 N. There is also confusion regarding the appropriate equation to determine the acceleration of the alpha particles resulting from this force. Clarifications are sought on the calculations, particularly regarding squaring the distance and doubling the charge. The conversation emphasizes the importance of accurate application of Coulomb's law in nuclear physics.
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1. The Nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons) (a) What is the force between the two alpha particles when they are
5*10^-15 m apart, and (b) what will be the magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u


Equations:

Coulomb's law: (k*(q1)*(q2))/r^2


Attempt:

i did k* (1.6*10^-19)2/ (5*10^-15) for part A and did not get the right answer which is supposed to be 36.8 N

- could someone please tell me how i am supposed to get that answer

- also i am not sure what equation to use for part 2
 
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You need to square both the distance and double the charge.
 
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