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NEET UG CHEMISTRY MCQ: STRUCTURE OF ATOM

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An α\alpha-particles of kinetic energy of 5.4 MeV is projected towards gold nucleus. Calculate the distance of closest approach. (Atomic number of gold =79,1 eV=1.6×10−19 J=79,1 \mathrm{~eV}=1.6 \times 10^{-19} \mathrm{~J})
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Correct option: A
4.21×10−14 m4.21 \times 10^{-14} \mathrm{~m}
Option A (4.21×10−14 m4.21 \times 10^{-14} \mathrm{~m}) is correct because it is calculated using the conservation of energy, equating the initial kinetic energy to the potential energy at the distance of closest approach.

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