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katkin
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#1
Report Thread starter 16 years ago
#1
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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username9816
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#2
Report 16 years ago
#2
(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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