JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    For emission line of atomic hydrogen from \[{{n}_{1}}=8\] to \[{{n}_{f}}=n\], the plot of wave number \[\left( \overline{v} \right)\] against \[\left( \frac{1}{{{n}^{2}}} \right)\] will be (The Rydberg constant, \[{{R}_{H}}\] is wave number unit) [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) Linear with slope - \[{{R}_{H}}\]

    B) Linear with slope -\[{{R}_{H}}\]

    C) Non linear

    D) Linear with intercept - \[{{R}_{H}}\]

    Correct Answer: B

    Solution :

       \[\frac{1}{\lambda }\,\,=\,\,{{R}_{H}}\,\left( \frac{1}{{{n}^{2}}_{1}}\frac{1}{{{n}^{2}}_{2}} \right)\] \[\overline{v}\,\,=\,\,{{R}_{H}}\,\left( \frac{1}{{{n}^{2}}}\,\,-\,\,\frac{1}{{{8}^{2}}} \right)\] \[\therefore \] Slope of the curve \[(\overline{v}\,\,Vs\,\,\frac{1}{{{n}^{2}}})\] is + RH Answer should be 2 but NTA has given answer 1. Which is not correct.


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