In the Rutherford scattering experiment, an a particle is | StudySoup

Textbook Solutions for Chemistry

Chapter 2 Problem 116P

Question

In the Rutherford scattering experiment, an \(\alpha\) particle is heading directly toward a gold nucleus. The particle will come to a halt when its kinetic energy is completely converted to electrical potential energy. When this happens, how close will the \(\alpha\) particle with a kinetic energy of \(6.0\times10^{-14}\mathrm{\ J}\) be from the nucleus? [According to Coulomb's law, the electrical potential energy between two charged particles is \(E=k Q_{1} Q_{2} / r\), where \(Q_{1}\) and \(Q_{2}\) are the charges in coulombs) of the \(\alpha\) particle and the gold nucleus, \(r\) is the distance of separation in meters, and \(k\) is a constant equal to \(9.0\times10^9\mathrm{\ kg}\cdot\mathrm{m}^3/\mathrm{s}^2\cdot\mathrm{C}^2\). Joule (\(\mathrm{J}\)) is the unit of energy where \(1\mathrm{\ J}=1\mathrm{\ kg}\cdot\mathrm{m}^2/\mathrm{s}^2\).]

Solution

Step 1 of 2

The goal of the problem is to find how close the  particle with a kinetic energy of  is from the nucleus.

 

Given:

According to Coulomb's law, the electrical potential energy between two charged particles is given by

,

where Q1 and Q2 are the charges (in coulombs) of the a particle and the gold nucleus.

 r = distance of separation in meters, and

k = constant equal to  ·

 

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full solution

Title Chemistry 11 
Author Raymond Chang
ISBN 9780073402680

In the Rutherford scattering experiment, an a particle is

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