JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Graphical Questions

  • question_answer
    The log-log graph between the energy E of an electron and its de-Broglie wavelength \[\lambda \] will be

    A)                                    

    B)                            

    C)                                    

    D)

    Correct Answer: C

    Solution :

                       \[\lambda =\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2m}}\cdot \frac{1}{\sqrt{E}}\]. Taking log of both sides \[\log \lambda =\log \frac{h}{\sqrt{2}m}+\log \frac{1}{\sqrt{E}}\]Þ\[\log \lambda =\log \frac{h}{\sqrt{2}m}-\frac{1}{2}\log E\]            \[\Rightarrow \log \lambda =-\frac{1}{2}\log E+\log \frac{h}{\sqrt{2m}}\]            This is the equation of straight line having slope (?1/2) and positive intercept on log l axis.


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