JCECE Medical JCECE Medical Solved Paper-2010

  • question_answer
    The wavelength of radiation emitted is \[{{\lambda }_{0}}\] when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to-the second orbit of hydrogen atom, the wavelength of radiation emitted will be

    A) \[\frac{16}{25}{{\lambda }_{0}}\]

    B)  \[\frac{20}{27}{{\lambda }_{0}}\]

    C)  \[\frac{27}{20}{{\lambda }_{0}}\]

    D)  \[\frac{25}{16}{{\lambda }_{0}}\]

    Correct Answer: B

    Solution :

     Wavelength \[(\lambda )\] during transition from \[{{n}_{2}}\]to\[{{n}_{1}}\] is given by \[\frac{1}{\lambda }=R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[\Rightarrow \] \[\frac{1}{{{\lambda }_{3\to 2}}}=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]=\frac{5R}{36}\] and \[\frac{1}{{{\lambda }_{4\to 2}}}=r\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]=\frac{3R}{16}\] \[\therefore \] \[\frac{{{\lambda }_{4\to 2}}}{{{\lambda }_{3\to 2}}}=\frac{20}{27}\] \[\Rightarrow \] \[{{\lambda }_{4\to 2}}=\frac{20}{7}{{\lambda }_{0}}\]


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