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An α-particle of mass m approaches a very heavy nucleus of charge Ze with speed v. The minimum distance it can get to the nucleus is proportional to which power of m?

1. 1/m
2. 1/√m
3. 1/m²
4. m

Correct answer: 1/m

Solution

At closest approach all KE converts to potential energy: (1/2)m v^2 = k(2e)(Ze)/rmin, so rmin = 4kZe^2/(m v^2), i.e. rmin is proportional to 1/m.

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