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A proton of energy 1.6 MeV approaches a gold nucleus (Z = 79). Find the distance of its closest approach.

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\(r= \frac{1}{4 \pi \epsilon _0} \frac{ze^2}{K.E.}\)

\(K.E. = 1.6 \times 10^6 \times 1.6 \times 10^{-19}\)

\(= 2.56 \times 10^{-13} J\)

\(r = \frac{9 \times 10^9\times 79\times (1.6\times 10^{-19})^2}{2.56 \times 10^{-13}}\)

\(= \frac{9 \times 10^9 \times 79 \times 1.6 \times 1.6 \times 10^{-38}}{2.56\times 10^{-13}}\)

\(= 9 \times 79 \times 10^{-16} = 711 \times 10^{-16} m\)

\(R \approx 71.1 fm\)

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