Atoms & Nuclei

JEE Main level, single correct, medium

7 of 239

Question 7 of 239

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=79 and 14πϵo=9×109\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} in SI units)

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Answer: (D) 2.95×10−142.95 \times 10^{-14}

At closest approach, KE converts to PE:
KE=14πε0×(2e)(Ze)rminKE = \frac{1}{4\pi\varepsilon_0} \times \frac{(2e)(Ze)}{r_{min}}
rmin=9×109×2×79×(1.6×10−19)27.7×106×1.6×10−19r_{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}}
=9×2×79×1.6×10−197.7×103= \frac{9 \times 2 \times 79 \times 1.6 \times 10^{-19}}{7.7} \times 10^{3}
=2275.27.7×10−16=2.95×10−14= \frac{2275.2}{7.7} \times 10^{-16} = 2.95 \times 10^{-14} m

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