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JEE Main 2026 (21 Jan, Shift 1) · Atomic Models & Alpha-Particle Scattering

Atomic Models & Alpha-Particle Scattering question from JEE Main 2026 (21 Jan, Shift 1)

Q4JEE Main 2026 · 21 JAN · SHIFT 1Medium

Atomic Models & Alpha-Particle Scattering

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 14πϵ0=9×109\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} in SI units)

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Answer: (c)

  1. r0=k(2e)(Ze)Kr_0 = \dfrac{k(2e)(Ze)}{K}, with K=7.7×106eK = 7.7\times10^6 e J.
  2. r0=9×109×2×79×1.6×10−197.7×106r_0 = \dfrac{9\times10^9 \times 2 \times 79 \times 1.6\times10^{-19}}{7.7\times10^6}
  3. r0=2.275×10−77.7×106≈2.95×10−14r_0 = \dfrac{2.275\times10^{-7}}{7.7\times10^6} \approx 2.95\times10^{-14} m.
  4. So the answer is (c).
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