In the First Nuclear Reaction Observed, Physicists Saw an Alpha $( Z = 2 )$

question 103

Short Answer

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 $55 \mathrm { MeV }$ , 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. $\left( 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} , e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , 1 / 4 \pi \varepsilon _ { 0 } = 8.99 \times 10 ^ { 9 } \mathrm {~N} \right.$ $\cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 }$ )