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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    One of the lines in the emission spectrum of \[L{{i}^{2+}}\] has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is \[=12\to n=x.\]. Find the value of x.

    A) \[8\]

    B)        \[6\]

    C) \[7\]

    D)        \[5\]

    Correct Answer: B

    Solution :

    For 2nd line of Balmer series in hydrogen spectrum \[\frac{1}{\lambda }=R(1)\,\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3}{16}R\] For \[L{{i}^{2+}}\left[ \frac{1}{\lambda }=R\times 9\left( \frac{1}{{{x}^{2}}}-\frac{1}{{{12}^{2}}} \right)=\frac{3R}{16} \right]\] which is satisfied by \[n=12\to n=6.\]


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