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JCECE Medical JCECE Medical Solved Paper-2010

  • question_answer
    Radius of \[{{\,}_{\text{2}}}\text{H}{{\text{e}}^{\text{4}}}\]nucleus is 3 fermi. The radius of \[_{\,\text{82}}\text{P}{{\text{b}}^{\text{206}}}\]nucleus will be

    A)  5 fermi         

    B)  6 fermi

    C)  11.16 fermi      

    D)  8 fermi

    Correct Answer: C

    Solution :

     We have, \[r\propto {{A}^{1/3}}\] \[\Rightarrow \] \[\frac{{{r}_{2}}}{{{r}_{1}}}={{\left[ \frac{{{A}_{2}}}{{{A}_{1}}} \right]}^{1/3}}\] \[={{\left[ \frac{206}{4} \right]}^{1/3}}\] \[\ \therefore \] \[{{r}_{2}}=3{{\left[ \frac{206}{4} \right]}^{1/3}}=11.16\,\text{fermi}\]


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