An alpha particle with a kinetic energy of 10.0 MeV makes a head-on collision with a gold nucleu

Name: An alpha particle with a kinetic energy of 10.0 MeV makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is 79, and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)
Uploaded: 2022-06-02T14:19:50-08:00
Duration: 1 min 27 s
Channel: Khoobchandra Agrawal
Description: An alpha particle with a kinetic energy of 10.0 MeV makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is 79, and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)

Step 1: Calculate the potential energy at the closest approach using the formula for potential e

Question

An alpha particle with a kinetic energy of $10.0 \mathrm{MeV}$ makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is $79,$ and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)

          An alpha particle with a kinetic energy of $10.0 \mathrm{MeV}$ makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is $79,$ and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)

Added by Rocio H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Khoobchandra Agrawal

06/02/2022

Step 1

99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)) $ q_1 $ = charge of alpha particle ($ 2e $) $ q_2 $ = charge of gold nucleus ($ 79e $) $ r $ = distance of closest approach Show more…

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An alpha particle with a kinetic energy of $10.0 \mathrm{MeV}$ makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is $79,$ and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)