If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is \_\_\_\_ m. (Atomic number of gold =79 and 1 4 π _ o =9 × 10^ 9 in SI units)

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Correct answer: D. 2.95 × 10^ -14

At closest approach, KE converts to PE:

KE = \frac{1}{4\pi\varepsilon_0} \times \frac{(2e)(Ze)}{r_\min}

r_\min = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}

= \frac{9 \times 2 \times 79 \times 1.6 \times 10^{-19}}{7.7} \times 10^{3}

= \frac{2275.2}{7.7} \times 10^{-16} = 2.95 \times 10^{-14}

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If an alpha particle with energy 7.7 MeV is bombarded on a thin...

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