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    Can anyone help me with this question?

    How close would a 6MeV alpha particle come to an Au nucleus in a head on collision, given that at a distance,r, it will stop and have an electrical PE of:
    PE=q1q2
    4piE.r

    Au nucleus charge = 79e-
    alpha particle charge = 2e-
    1 = k=9*10 to the 9 (Nm squared c to the -2)
    4piE.
    E. = 8.85 * 10 to the -12 (c squared N to the -1 m to the -2)
    lev?= energy = i.6 * 10 to the -19 J
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    (Original post by katkin)
    Can anyone help me with this question?

    How close would a 6MeV alpha particle come to an Au nucleus in a head on collision, given that at a distance,r, it will stop and have an electrical PE of:
    PE=q1q2
    4piE.r

    Au nucleus charge = 79e-
    alpha particle charge = 2e-
    1 = k=9*10 to the 9 (Nm squared c to the -2)
    4piE.
    E. = 8.85 * 10 to the -12 (c squared N to the -1 m to the -2)
    lev?= energy = i.6 * 10 to the -19 J
    :confused:
 
 
 
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