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    Rutherford observion of an alpha particle

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    Rutherford observed that an alpha particle (Q1 = 2 x 1.6 x 10^-19 coulomb) having a kinetic energy of 7.68 x 10 ^6 electron volts (7.68 x 10^6 x 1.6 x 10^-19 joule) rebounds backward in a head on collision with a gold nucleus (Q2 = 79 x 1.6 x 10^-19 coulomb).
    a. What is the distance of closest approach where the electrostatic potential energy is equal to the initial kinetic energy? Express results in femtometers (10^-15 meter).
    b. What is the maximum force of repulsion?
    c. What is the maximum acceleration in g's? The mass of the alpha particle is about 4 times that of a proton, or 4 x 1.7 x 10^-27 kilogram.

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    Rutherford observed that an alpha particle (Q1 = 2 x 1.6 x 10-19C) having a kinetic energy of 7.68 x 106 eV (7.68 x 106 x 1.6 x 10-19J) rebounds ...

    Solution Summary

    The solution calculates the distance of the closest approach where the electrostatic potential energy is equal to the initial kinetic energy. The maximum force of repulsion and maximum acceleration is calculated.

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