CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    The wave number of the spectral line in the emission spectrum of hydrogen will be equal to \[\frac{8}{9}\] times the Rydbergs constant if the electron jumps from

    A)  \[n=3\] to \[n=1\] 

    B)  \[n=10\] to \[n=1\]

    C)  \[n=9\] to \[n=1\]

    D)  \[n=2\] to \[n=1\]

    Correct Answer: A

    Solution :

    Wave number of spectral line in emission spectrum of hydrogen, \[\overline{v}={{R}_{H}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\]            ??.(i) Given, \[\overline{v}=\frac{8}{9}{{R}_{H}}\] On putting the value of v in Eq. (i), we get \[\frac{8}{9}{{R}_{H}}={{R}_{H}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\]          \[\frac{8}{9}-1=-\frac{1}{{{n}_{2}}}\] \[\frac{1}{3}=\frac{1}{{{n}_{2}}}\] \[\therefore \] \[{{n}_{2}}=3\] Hence, electron jumps from \[{{n}_{2}}=3\] to \[{{n}_{1}}=1\]


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