BCECE Medical BCECE Medical Solved Papers-2014

  • 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                 \[\frac{8}{9}{{R}_{H}}={{R}_{H}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\]                 \[\frac{8}{9}=\frac{1}{{{(1)}^{2}}}-\frac{1}{n_{2}^{2}}\]                 \[\frac{1}{3}-1=-\frac{1}{n_{2}^{2}}\,\,\,\therefore \,\,\,{{n}_{2}}=3\] Hence, electron jumps from \[{{n}_{2}}=3\] to \[{{n}_{1}}=1\]


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