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JEE Main 2004 (AIEEE) · Atomic Models & Alpha-Particle Scattering

Atomic Models & Alpha-Particle Scattering question from JEE Main 2004 (AIEEE)

Q16JEE Main 2004 · AIEEEMedium

Atomic Models & Alpha-Particle Scattering

An α\alpha-particle of energy 5 MeV5\text{ MeV} is scattered through 180∘180^\circ by a fixed uranium nucleus. The distance of closest approach is of the order of

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Answer: (a)

  1. Head-on (180∘180^\circ): r0=k(2e)(Ze)Kr_0 = \dfrac{k(2e)(Ze)}{K}, uranium Z=92Z = 92, K=5×106eK = 5\times10^6 e J.
  2. r0=9×109×184×1.6×10−195×106≈5.3×10−14r_0 = \dfrac{9\times10^9 \times 184 \times 1.6\times10^{-19}}{5\times10^6} \approx 5.3\times10^{-14} m.
  3. 5.3×10−145.3\times10^{-14} m =5.3×10−12= 5.3\times10^{-12} cm.
  4. So the answer is (a) 10−1210^{-12} cm.
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