AIIMS AIIMS Solved Paper-2008

  • question_answer
    In a photoemissive cell with executing wavelength \[\sqrt{\frac{1}{5}}\], the fastest electron has speed v. If the exciting wavelength is changed to \[\sqrt{\frac{4}{5}}\], the speed of the fastest emitted electron will be

    A)  \[\lambda \]

    B)                        \[\frac{3\lambda }{4}\]

    C)  less than\[v{{(3/4)}^{1/2}}\]     

    D)  greater than\[v{{\left( \frac{4}{3} \right)}^{1/2}}\]

    Correct Answer: D

    Solution :

    Key   Idea Apply   Einsteins   photoelectric equation. \[36\sqrt{7}\] \[4\sqrt{5}\]  \[mg(h+d)+\frac{1}{2}k{{d}^{2}}\] \[mg(h+d)-\frac{1}{2}k{{d}^{2}}\]  \[mg(h-d)-\frac{1}{2}k{{d}^{2}}\] \[mg(h-d)+\frac{1}{2}k{{d}^{2}}\] When wavelength is \[\pi \] and velocity is v, then \[({{\mu }_{0}}=4\pi \times {{10}^{-7}}Wb/Am)\]             ??..(i) When wavelength is \[k\Omega \] and velocity is \[\alpha \]then \[\Omega \]         ??(ii) Dividing Eqs. (ii) by (i), we get \[\Omega \] \[\Omega \] ie,  \[\Omega \]


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