JCECE Engineering JCECE Engineering 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        \[\frac{1}{{{\lambda }_{4\to 2}}}=r\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]=\frac{3R}{16}\] \[\therefore \]  \[\frac{{{\lambda }_{4\to 2}}}{{{\lambda }_{3\to 2}}}=\frac{20}{27}\] \[\Rightarrow \]               \[{{\lambda }_{4\to 2}}=\frac{20}{27}{{\lambda }_{0}}\]


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