An alpha particle with kinetic energy 14.0 MeV makes a collision with a 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=(p0)b, where p0 is the magnitude of the initial linear momentum of the alpha particle and b = 1.40×10^-12 m. (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.)
What is the distance of closest approach?
Repeat for b = 1.40×10^-13 m.
Repeat for b = 1.30×10^-14 m.© BrainMass Inc. brainmass.com October 24, 2018, 11:41 pm ad1c9bdddf
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