MGIMS WARDHA MGIMS WARDHA Solved Paper-2004

  • question_answer
    If work function of a metal is 4.2 eV, the cut off wave length is :

    A) \[8000\overset{o}{\mathop{\text{A}}}\,\]

    B) \[7000\overset{o}{\mathop{\text{A}}}\,\]

    C) \[1472\overset{o}{\mathop{\text{A}}}\,\]

    D) \[2950\overset{o}{\mathop{\text{A}}}\,\]

    Correct Answer: D

    Solution :

    Suppose the cut off wavelength is represented by\[{{\lambda }_{0}}\]. so,\[\frac{hc}{{{\lambda }_{0}}}=\]work function\[={{W}_{0}}\] \[\frac{hc}{{{\lambda }_{0}}}=4.2\times 1.6\times {{10}^{-19}}\]                 \[{{\lambda }_{0}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4.2\times 1.6\times {{10}^{-19}}}\]                 \[=2946\times {{10}^{-10}}m\]                 \[\approx 2950{\AA}\]


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