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An alpha particle with kinetic energy 14.0 MeV makes a collision with lead nucleus, but it...

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Physics

An alpha particle with kinetic energy 14.0 MeV makes a collisionwith lead nucleus, but it is not \"aimed\" at the center of the leadnucleus, and has an initial nonzero angular momentum (with respectto the stationary lead nucleus) of magnitude L=p0b, where p0 is themagnitude of the initial momentum of the alpha particle andb=1.50×10−12 m . (Assume that the lead nucleus remains stationaryand that it may be treated as a point charge. The atomic number oflead is 82. The alpha particle is a helium nucleus, with atomicnumber 2.)

A) What is the distance of closest approach?

B) Repeat for b=1.50×10−13 m .

C) Repeat for b=1.50×10−14 m .

Answer & Explanation Solved by verified expert
3.6 Ratings (356 Votes)
Answer The total energy at closest approach is the kinetic energy potential energy at that point Let v0 be the velocity and r0 the distance at the    See Answer
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