Explanation If recoil motion is ignored, at the closest approach the kinetic energy of both particles is zero. The potential energy of the two charges must equal the kinetic energy of the α particle. K = ( 1 4 π ε 0 ) ( Z α e ) ( Z Ag e ) r min r min = ( 1 4 π ε 0 ) ( Z α ) ( Z Ag ) e 2 K Here, K is the kinetic energy, r min is the minimum radius, Z α is the atomic number of the α particle, e is the charge of the electron and Z Ag is the atomic number of silver. Conclusion: Substitute 9.00 × 10 9 N ⋅ m 2 / C 2 for ( 1 4 π ε 0 ) , 2 for Z α , 47 for Z Ag , 1

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Chapter 37, Problem 81GP

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The closeness of α particles to the center of a silver nucleus.

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Chapter 37, Problem 80GP

Chapter 37, Problem 82GP

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Chapter 37 Solutions

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