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JEE Main & Advanced JEE Main Paper (Held On 12 May 2012)

  • question_answer
    A doubly ionised Li atom is excited from its ground state (n=1) to n=3 state. The wavelengths of the spectral lines are given by \[{{\lambda }_{32}},{{\lambda }_{31}}\]and\[{{\lambda }_{21}}.\]The ratio \[{{\lambda }_{32}}/{{\lambda }_{31}}\]and\[{{\lambda }_{21}}/{{\lambda }_{31}}\]are, respectively     JEE Main Online Paper (Held On 12 May 2012)

    A) 8.1,0.67               

    B)                        8.1,1.2

    C)                        6.4,1.2                  

    D)                        6.4,0.67

    Correct Answer: C

    Solution :

                    \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\]where R = Rydberg constant\[\frac{1}{{{\lambda }_{32}}}=\left( \frac{1}{4}-\frac{1}{9} \right)=\frac{5}{36}\]\[\Rightarrow \]\[{{\lambda }_{32}}=\frac{36}{5}\] Similarly solving for \[{{\lambda }_{31}}\]and \[{{\lambda }_{21}}\] \[{{\lambda }_{31}}=\frac{9}{8}\]and \[{{\lambda }_{21}}=\frac{4}{3}\] \[\therefore \]\[\frac{{{\lambda }_{32}}}{{{\lambda }_{31}}}=6.4\]and\[\frac{{{\lambda }_{21}}}{{{\lambda }_{31}}}\simeq 1.2\]


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