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

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In the first nuclear reaction observed, physicists saw an alpha particle (Z=2)( Z = 2 ) interact with a nitrogen nucleus in air (Z=7)( Z = 7 ) to produce a proton. The energy of the alpha particle was 55MeV55 \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. (1eV=1.60×1019 J,e=1.60×1019C,1/4πε0=8.99×109 N\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. m2/C2\cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } )


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