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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    The radius of hydrogen atom is\[0.53\overset{o}{\mathop{\text{A}}}\,\]in the ground state. The radius of\[L{{i}^{2+}}\]ion\[(Z=3)\] in this state is

    A)  \[0.17\overset{o}{\mathop{\text{A}}}\,\]           

    B)  \[1.06\overset{o}{\mathop{\text{A}}}\,\]

    C)  \[0.53\overset{o}{\mathop{\text{A}}}\,\]            

    D)  \[0.265\overset{o}{\mathop{\text{A}}}\,\]

    Correct Answer: A

    Solution :

     Atomic radius\[=\frac{0.53{{n}^{2}}}{Z}\overset{o}{\mathop{\text{A}}}\,\] where   \[n=\]number of orbital \[Z=\]atomic number radius or\[L{{i}^{2+}}\]ion\[=\frac{0.53\times 1}{3}\overset{o}{\mathop{\text{A}}}\,\] \[=0.176\overset{o}{\mathop{\text{A}}}\,\]


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