RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    The wavelength of second line of Lyman series is 1025\[\overset{o}{\mathop{\text{A}}}\,\]. The wavelength of first line will be

    A)  825\[\overset{o}{\mathop{\text{A}}}\,\]

    B)  900\[\overset{o}{\mathop{\text{A}}}\,\]

    C)  1215\[\overset{o}{\mathop{\text{A}}}\,\]          

    D)  1325\[\overset{o}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

     \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{\left( 1-\frac{1}{9} \right)}{\left( 1-\frac{1}{4} \right)}\] Or \[{{\lambda }_{1}}=\frac{8\times 4}{9\times 3}\times 1025\] \[{{\lambda }_{1}}=1215\overset{o}{\mathop{\text{A}}}\,\]


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