Explanation Given Info : The charge on the alpha particle is equal to + 2 e . The charge on the gold atom is equal to + 79 e . The initial kinetic energy of the alpha particle is K . The value of K is 5.3 MeV . In Rutherford’s scattering experiments, alpha particles were fired at a gold foil. The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. The expression for the kinetic energy to calculate the distance of closest approach is, K = q 1 × q 2 4 π ε ο r (1) Here, q 1 is the charge on the alpha particle. q 2 is the charge on the gold atom. ε ο is the permittivity of the free space. its value is 8.85 × 10 − 12 C 2 /N ⋅ m 2 . r is the radius. To calculate the value of distance of the closest approach between the alpha particle and the gold nucleus, the equation (1) is modified. The value of K in joule unit is 5.3 × 1.6 × 10 − 13 J . The formula to calculate the value of distance of the closest approach between the alpha particle and the gold nucleus is, r = q 1 × q 2 4 π ε ο K Substitute 8.85 × 10 − 12 C 2 /N ⋅ m 2 for ε ο , + 2 e for q 1 , + 79 e for q 1 , 1.6 × 10 − 19 C for e , 5

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ISBN: 9780321976444

Author: James S. Walker

Publisher: PEARSON

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Rutherford's scattering experiment

Rutherford’s scattering experiment is also known as the gold foil experiment. This experiment disproved the plum pudding model of J.J. Thomson. In a gold foil experiment, high-energy alpha particles are made to strike a thin gold foil. The scattered alph…

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Chapter 31, Problem 5PCE

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The distance of closest approach between the alpha particle and the gold nucleus for the case K=5.3 MeV .

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Chapter 31, Problem 4PCE

Chapter 31, Problem 6PCE

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The distance of closest approach between the alpha particle and the gold nucleus for the case K = 5.3 MeV .

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The distance of closest approach between the alpha particle and the gold nucleus for the case K=5.3 MeV .

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Chapter 31 Solutions