JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Critical Thinking

  • question_answer
    \[\alpha -\]particles of energy 400 KeV are bombarded on nucleus of\[_{82}Pb\]. In scattering of \[\alpha -\]particles, its minimum distance from nucleus will be                                                                 [RPET 1997]

    A)            0.59 nm                                  

    B)            0.59 Å

    C)            5.9 pm                                    

    D)            0.59 pm

    Correct Answer: D

    Solution :

               Suppose closest distance is r, according to conservation of energy.                    \[400\times {{10}^{3}}\times 1.6\times {{10}^{-19}}=9\times {{10}^{9}}\frac{(ze)\,(2e)}{r}\]                    \[\Rightarrow 6.4\times {{10}^{-14}}\]\[=\frac{9\times {{10}^{9}}\times (82\times 1.6\times {{10}^{-19}})\times (2\times 1.6\times {{10}^{-19}})}{r}\]                    \[\Rightarrow r=5.9\times {{10}^{-13}}m=0.59\,pm\].


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