An alpha particle with kinetic energy 9.50 MeV (when far away) collides head-on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and 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.)
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A: m1=1.95g=0.00195Kgq1=15.0μC=15×10−6Cv1i=16m/sm2=5.25g=0.00525Kgq2=8.50μC=8.50×10−6C
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An alpha particle with kinetic energy 9.50 MeV (when far
away) collides head-on with a lead nucleus at rest. What is the distance
of closest approach of the two particles? (Assume that the lead
nucleus remains stationary and 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.)
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