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NEET Physics Atomic Physics NEET PYQ-Atomic Physics

  • question_answer
    An alpha nucleus of energy \[\frac{1}{2}m{{v}^{2}}\] bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

    A)  \[\frac{1}{Ze}\]            

    B)  \[{{v}^{2}}\]

    C)  \[\frac{1}{m}\]            

    D)       \[\frac{1}{{{v}^{4}}}\]

    Correct Answer: B

    Solution :

    An \[\alpha \]-particle of mass m possesses initial velocity v, when it is at a large distance from the nucleus of an atom having atomic number Z. At the distance of closest approach, the kinetic energy of \[\alpha \]-particle is completely converted into potential energy. Mathematically,
    \[\frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(2e)(Ze)}{{{r}_{0}}}\]
    \[{{r}_{0}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2Z{{e}^{2}}}{\frac{1}{2}m{{v}^{2}}}\]


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