The first nuclear reaction observed saw an alpha particle (Z = 2)interact with a nitrogen nucleus in air (Z = 7)to produce a proton.The energy of the alpha particle was 55 MeV,enough to enable the nuclei to touch in spite of the Coulomb repulsion.What distance between the centers of the alpha particle and the nitrogen was reached? This determines a nuclear radius for nitrogen.Use 1 fm = 10^-15 m.

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Quizplus · Accepted Answer

The distance of closest approach can be calculated using the formula for the Coulomb barrier: $ r = \frac{k \cdot e^2 \cdot Z_1 \cdot Z_2}{E} $, where $ k $ is Coulomb's constant, $ e $ is the elementary charge, $ Z_1 $ and $ Z_2 $ are the atomic numbers, and $ E $ is the energy. Substituting the given values, the closest approach is approximately 0.37 fm.

The First Nuclear Reaction Observed Saw an Alpha Particle (Z