Uttarakhand PMT Uttarakhand PMT Solved Paper-2006

  • question_answer
    If work function of a metal is 4.2eV, the cut off wavelength 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 \[={{W}_{0}}\] \[\frac{hc}{{{\lambda }_{0}}}=4.2\times 1.6\times {{10}^{-19}}\] \[{{\lambda }_{0}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4.2\times 1.6\times {{10}^{-19}}}\] \[=2946\times {{10}^{-10}}m\] \[\approx 2950\overset{o}{\mathop{\text{A}}}\,\]


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