RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    The wavelength of first line of balmer series is 6563\[\overset{o}{\mathop{\text{A}}}\,\]. The wavelength of first line of lymen series will be

    A)  1215.4\[\overset{o}{\mathop{\text{A}}}\,\]         

    B)  2500\[\overset{o}{\mathop{\text{A}}}\,\]

    C)  7500\[\overset{o}{\mathop{\text{A}}}\,\]            

    D)  600\[\overset{o}{\mathop{\text{A}}}\,\]

    Correct Answer: A

    Solution :

     \[\frac{{{\lambda }_{Lymen}}}{{{\lambda }_{Balmer}}}=\frac{\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)}{\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)}=\frac{5}{27}\] \[\left[ \because \frac{1}{\lambda }=R\left( \frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}} \right) \right]\] \[{{\lambda }_{Lymen}}=\frac{5}{27}\times {{\lambda }_{Bamer}}\] \[=\frac{5}{27}\times 6563=1215.4\overset{o}{\mathop{\text{A}}}\,\]


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