RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    Wavelength of second line of Lyman series is \[1025\overset{\text{o}}{\mathop{\text{A}}}\,,\] then wavelength of first line is:

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

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

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

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

    Correct Answer: C

    Solution :

                    For Lyman series, \[\frac{1}{\lambda }=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{n}^{2}}} \right)\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right)}{\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)}\] \[n=2,3,4.......\] For 1st line \[n=2\], for IInd line \[n=3\]                 \[{{\lambda }_{1}}=1215\overset{\text{o}}{\mathop{\text{A}}}\,\]


You need to login to perform this action.
You will be redirected in 3 sec spinner