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An alpha particle with kinetic energy 15.

This question was answered on Sep 22, 2013. View the Answer
An alpha particle with kinetic energy 15.0MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p_0b, where p_0 is the magnitude of the initial momentum of the alpha particle and b=1.40×10^−12m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

Find distance of closest approach r when
a) b=1.50x10^-13m
b) b=1.30x10^-14m
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This question was asked on Sep 21, 2013 and answered on Sep 22, 2013.

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