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, 1/4??₀ = 8.99 × 10⁹ N ? m²/C²)

Question

Quizplus · Accepted Answer

The alpha particle and the nitrogen nucleus are both positively charged, so they repel each other. The energy of the alpha particle is not enough to overcome this repulsion, so the two nuclei cannot touch. The closest they can get is when their centers are separated by a distance of 0 fm.

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 $55 \mathrm { MeV } \text {, }$