Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    If levels 1 and 2 are separated by an energy \[C\xrightarrow[{}]{{}}D;{{k}_{2}}={{10}^{12}}{{e}^{-24,606/T}}\] such that the  corresponding transition frequency falls in the middle of the visible range, calculate the ratio of the populations of two levels in the thermal equilibrium at room temperature.

    A) \[{{k}_{1}}\]      

    B) \[{{k}_{2}}\]

    C) \[CaC{{l}_{2}}\]                

    D) \[C{{s}^{+}}\]

    Correct Answer: A

    Solution :

    At thermal equilibrium, the ratio \[y=\pm \sqrt{{{\log }_{e}}x{{)}^{2}}+2C}\] is given as\[xy={{x}^{y}}+C\] The middle of the visible range is taken at \[\frac{4}{{{100}^{3}}}\] \[\frac{3}{{{50}^{3}}}\]\[\frac{3!}{{{100}^{3}}}\] \[f(x)=\left\{ \begin{matrix}    \frac{\sin (\cos x)-\cos x}{{{(\pi -2x)}^{3}}}, & x\ne \frac{\pi }{2}  \\    k, & x=\frac{\pi }{2}  \\ \end{matrix} \right.\]\[x=\frac{\pi }{2},\] \[-\frac{1}{6}\]\[-\frac{1}{24}\]


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