Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    In a photo emissive cell with exciting wavelength \[{{\tan }^{-1}}\left( \frac{{{n}^{2}}-n}{{{n}^{2}}-n+2} \right)\]the fastest electron has speed \[{{\tan }^{-1}}\left( \frac{{{n}^{2}}+n+2}{{{n}^{2}}+n} \right)\]. If the exciting wavelength is changed by \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\left( 1-\tan \frac{x}{2} \right)(1-\sin x)}{\left( 1+\tan \frac{x}{2} \right){{(\pi -2x)}^{3}}}\], the speed of the fastest emitted electron will be

    A) \[\frac{1}{8}\]

    B) \[\frac{1}{32}\]

    C) less than \[\infty \]        

    D) greater than\[f(x)=\left\{ \begin{align}   & \frac{{{\sin }^{3}}(\sqrt{3}).\log (1+3x)}{{{({{\tan }^{-1}}\sqrt{x})}^{2}}({{e}^{5\sqrt{x}}}-1)x},x\ne 0 \\  & a,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=0 \\ \end{align} \right.\]

    Correct Answer: D

    Solution :

    From\[5.00\pm 0.10cm\]       (where \[{{m}_{A}}\]) If wavelength of incident light charges from \[{{r}_{A}}\]to \[{{r}_{B}}\](decreases). Let. energy of incident light charges from E to E' and speed of fastest electron changes from v to Ö then \[{{T}_{A}}\]                       ?(i) andÖ\[{{T}_{B}}\]                                             ?(ii) As\[{{m}_{A}}\] Hence, Ö\[{{m}_{B}}\] \[\frac{{{T}_{A}}}{{{T}_{B}}}={{\left( \frac{{{r}_{A}}}{{{r}_{B}}} \right)}^{3/2}}\]Ö\[{{T}_{A}}>{{T}_{B}}(if\,{{r}_{A}}>{{r}_{B}})\] \[{{T}_{A}}>{{T}_{B}}(if\,{{m}_{A}}>{{m}_{B}})\]Ö\[{{T}_{A}}={{T}_{B}}\] So Ö\[\frac{2R}{\sqrt{15}}\]


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