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Atoms and Nuclei question

2010 · Q159
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Atoms and Nuclei question

2010 · Q159

NEETPhysicsAtoms and NucleiMCQ+4 / −1
An alpha nucleus of energy 12{1 \over 2}21​ mv2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
  1. A
    1Ze{1 \over {Ze}}Ze1​
  2. B
    v2
  3. C
    1m{1 \over m}m1​
  4. D
    1v4{1 \over {{v_4}}}v4​1​
View written solutionFree

Correct answer: C

Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,

12mv2=14πε0×(Ze)×(2e)din{1 \over 2}m{v^2} = {1 \over {4\pi {\varepsilon _0}}} \times {{\left( {Ze} \right) \times \left( {2e} \right)} \over {{d_{in}}}}21​mv2=4πε0​1​×din​(Ze)×(2e)​

∴\therefore∴ dmin ∝\propto∝ 1m{1 \over m}m1​

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