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AMU Medical AMU Solved Paper-2010

  • question_answer
    Light of two different frequencies, whose photons have energies 1 eV and 2.5 eV are respectively, successively illuminate a surface of work function 0.5 eV. The ratio of the maximum speeds of the electrons that are emitted is

    A)  2 : 1                                      

    B)  1 : 2

    C)  1 : 3                                      

    D)  1 : 4

    Correct Answer: B

    Solution :

                     The maximum kinetic energy of the emitted electron                                 \[K=hv-{{W}_{0}}\] where hv = photon energy                 \[{{W}_{0}}=\] work function Suppose the mass of electron is m, then in the case I                 \[\frac{1}{2}mv_{1}^{2}=1\,eV-0.5=0.5\,eV\]     ... (i) In case II                                 \[\frac{1}{2}mv_{2}^{2}=2.5\,eV\,-0.5\,eV=2.0\,eV\]                 \[\therefore \]  \[\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{0.5}{2.0}=\frac{1}{4}\]                                 \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1}{2}\]


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