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NEET AIPMT SOLVED PAPER SCREENING 2011

  • question_answer
    Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be

    A)  1 : 2                                      

    B)  1 : 1                      

    C)  1 : 5                      

    D)         1 : 4

    Correct Answer: A

    Solution :

    Kinetic energy                                    \[KE=\phi -{{\phi }_{0}}\] Here,     \[K{{E}_{1}}=1-0.5=0.5eV\] \[K{{E}_{2}}=2.5-0.5=2eV\] \[\therefore \]  \[\frac{K{{E}_{1}}}{K{{E}_{2}}}=\frac{0.5}{2}=\frac{1}{4}\] or            \[\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4}\]           or                \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\]


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