At closest approach, KE converts to PE:
KE=4πε01×rmin(2e)(Ze)
rmin=7.7×106×1.6×10−199×109×2×79×(1.6×10−19)2
=7.79×2×79×1.6×10−19×103
=7.72275.2×10−16=2.95×10−14 m
JEE Main 2026 — Physics Modern Physics
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 4πϵo1=9×109 in SI units)
Held on 21 Jan 2026 · Verified 6 Jul 2026.
3.85×10−16
3.85×10−14
2.95×10−16
2.95×10−14
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