RAJASTHAN PMT Rajasthan - PMT Solved Paper-2006

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

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

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

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

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

    Correct Answer: D

    Solution :

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


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