NEET AIPMT SOLVED PAPER SCREENING 2010

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

    A) \[\frac{\text{1}}{Ze}\]  

    B)                        \[{{\text{v}}^{\text{2}}}\]

    C) \[\frac{1}{m}\]                 

    D)        \[\frac{\text{1}}{{{\text{v}}^{\text{4}}}}\]

    Correct Answer: C

    Solution :

    An a-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 a-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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