
Concept explainers
The closest approach that a 5.3 MeV alpha particle can make to a gold nucleus.

Answer to Problem 3.1P
Explanation of Solution
Introduction:
When an alpha particle having kinetic energy ‘Ea’and charge ‘q’ is ejected on a gold nucleus, it undergoes repulsion. The force of repulsion increases with the decreasing distance between the two. On reaching the closest distance of approach ( d ), the total kinetic energy converts into potential energy hence, it comes to rest.
Formula:
Consider an alpha particle of energy
Using the formula, we get:
Conclusion:
Therefore, the closest distance of approach ( d ) for a 5.3 MeV alpha particle is:
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Chapter 3 Solutions
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