Haryana PMT Haryana PMT Solved Paper-2005

  • question_answer
    If the wavelength of the first line of the Balmer series of hydrogen is \[6561\overset{\text{o}}{\mathop{\text{A}}}\,\] the wavelength of second line of the series will be :

    A) \[13122\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B) \[2187\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) \[4860\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D) \[3280\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

                    For Balmer series \[{{n}_{1}}=2,{{n}_{2}}=3\] for 1st line and \[{{n}_{1}}=2,{{n}_{2}}=4\] for second line. Hence, \[\frac{1}{{{\lambda }_{1}}}=R\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]\]           ?..(i) \[\frac{1}{{{\lambda }_{2}}}=R\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(4)}^{2}}} \right]\]       ??..(ii) \[\therefore \]  \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{20}{27}\] or \[{{\lambda }_{2}}=\frac{20}{27}\times 6561=4860\overset{\text{o}}{\mathop{\text{A}}}\,\]


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