BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    Light of two different frequencies whose photons have energies 1 eV and 2.5 eV successively illuminate a metal of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons will be

    A)  1 : 5

    B)                         1 : 4

    C)  1 : 2                      

    D)         1 : 1

    Correct Answer: C

    Solution :

    According to Einsteins photoelectric equation, \[({{E}_{K}})=hv-{{\phi }_{0}}\] \[{{({{E}_{K}})}_{1}}=1-0.5=0.5\,eV\] Similarly, \[{{({{E}_{K}})}_{2}}=2.5-0.5=2\,eV\] \[\therefore \]  \[\frac{{{({{E}_{K}})}_{1}}}{{{({{E}_{K}})}_{2}}}=\frac{1}{4}\] \[\Rightarrow \]               \[\frac{\frac{1}{2}mv_{1}^{2}}{\frac{1}{2}mv_{2}^{2}}=\frac{1}{4}\] \[\Rightarrow \]               \[\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4}\] \[\Rightarrow \]               \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1}{2}\]


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