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Atoms and Nuclei question

2004 · Shift 0 · Q123
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Atoms and Nuclei question

2004 · Shift 0 · Q123

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
An α\alphaα-particle of energy 5MeV5MeV5MeV is scattered through 180∘{180^ \circ }180∘ by a fixed uranium nucleus. The distance of closest approach is of the order of
  1. A
    10−12 cm{10^{ - 12}}\,cm10−12cm
  2. B
    10−10 cm{10^{ - 10}}\,cm10−10cm
  3. C
    1A1A1A
  4. D
    10−15a cm{10^{ - 15a}}\,cm10−15acm
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