JCECE Engineering JCECE Engineering Solved Paper-2010

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

    A) \[5\]fermi                          

    B) \[6\]Fermi

    C) \[11.16\] fermi                 

    D)  \[8\] Fermi

    Correct Answer: C

    Solution :

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


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