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)

Question

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)

Quizplus · Accepted Answer

The Answer of 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)

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

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