In the First Nuclear Reaction Observed,physicists Saw an Alpha Particle $55 \mathrm { MeV } \text {, }$

In the first nuclear reaction observed,physicists 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 $55 \mathrm { MeV } \text {, }$ enough to enable the nuclei to touch in spite of the Coulomb repulsion.What distance (in fm) between the centers of the alpha particle and the nitrogen was reached? This determines a limit on the radius of the nitrogen nucleus.(1 eV = 1.60 × 10^-19 J,e = 1.60 × 10^-19 C,1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

A) 0.37 fm
B) 0.18 fm
C) 0.05 fm
D) 158.40 fm

Correct Answer:

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