Question
In: Physics
An alpha particle with kinetic energy 13.0 MeV makes a collision with lead nucleus, but it...
An alpha particle with kinetic energy 13.0 MeV 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=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.20×10?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=1.50×10?13 m .
-Repeat for b=1.30×10?14 m .
Solutions
genius_generous answered 2 years agoRelated Solutions
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magnitude of the initial momentum of the alpha particle and
b=1.20×10−12 m . (Assume that the lead nucleus remains stationary
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nucleus, and has an initial nonzero angular momentum (with respect
to the stationary lead nucleus) of magnitude L=p0b, where p0 is the
magnitude of the initial momentum of the alpha particle and
b=1.50×10−12 m . (Assume that the lead nucleus remains stationary
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An alpha particle with kinetic energy 14.0 MeV 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=p0b, where p0 is the
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b=1.20×10−12 m . (Assume that the lead nucleus
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