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Numerade Educator

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Problem 82 Hard Difficulty

An alpha particle with kinetic energy 11.0 MeV makes a head-on collision 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 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 $2.)$

Answer

$2.15 \times 10^{-14} \mathrm{m}$

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Video Transcript

here we have two particles that have the same charge and Cuban approaching Q two some kinetic energy given as 11 mega electron volts. And since they're both positively charged, there will be a repulsion. And assuming that, um, conservation of energy holds here, we can say the electric potential energy between the two charges, combined with the kinetic energy, will remain constant. So then we know that the electric potential Cornel was given by K Q one Q two over our So here we're trying to find what are is how close they get before they repulsed each other. I guess so, when key one is really far away from Q two are will be approaching infinity, meaning that the electric potential energy is negligible when Q one has approached you to, There will be a split second where the kinetic energy will be zero because Q one will not be moving. So then, here we have our 11. My goal left rumbles. Just given the problem, you could you okay, Q. One Q two over. Our I can rearrange us to get R is equal to okay, which is also sometimes written as 1/4 pi. Absolutely not underst keeping it simple here over 11 my girl Electron volts. And we've got our two charges up here and soon E is equal to 1.6 times 10 to the minus 19. We can then plug those in and getting our value equal to about 2.15 arms. Tongue tu minus four key meters.