EBK INTRODUCTION TO HEALTH PHYSICS, FIF
EBK INTRODUCTION TO HEALTH PHYSICS, FIF
5th Edition
ISBN: 9780071835268
Author: Johnson
Publisher: VST
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Chapter 3, Problem 3.1P
To determine

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

Expert Solution & Answer
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Answer to Problem 3.1P

  d=4.29×1014m

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:

  d=(14πε0)qQEa

Consider an alpha particle of energy Ea=5.3MeV=(5.3×106×1.6×1019)

Using the formula, we get:

  d=(14πε0)qQEa

  d=9×109(2×1.6× 10 19×79×1.6× 10 195.3×10×61.6× 10 19)

  d=4.29×1014m

Conclusion:

Therefore, the closest distance of approach ( d ) for a 5.3 MeV alpha particle is: d=4.29×1014m .

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