AMU Medical AMU Solved Paper-1999

  • question_answer Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength 2000 A falls on the surface, the velocity of the fastest electron ejected from the surface will be

    A) \[8.4\times {{10}^{5}}\]m/second

    B) \[7.4\times {{10}^{5}}\]m/second

    C) \[6.4\times {{10}^{5}}\]m/second

    D)  \[8.4\times {{10}^{6}}\]m/second

    Correct Answer: A

    Solution :

    : For photoelectric effect, \[\frac{hc}{\lambda }=\phi +\frac{1}{2}m{{v}^{2}}\]or\[\frac{1}{2}m{{v}^{2}}=\frac{hc}{\lambda }-\phi \] \[=\frac{(6.6\times {{10}^{-34}})\times (3\times {{10}^{8}})}{2000\times {{10}^{-10}}}-\frac{4.2\times 1.6\times {{10}^{-19}}}{1}\] Or \[\frac{1}{2}\times (9\times {{10}^{-31}})\times {{v}^{2}}\] \[=9.9\times {{10}^{-19}}-6.72\times {{10}^{-19}}\] \[=3.18\times {{10}^{-19}}\] Or \[{{v}^{2}}=\frac{2\times 3.18\times {{10}^{-19}}}{9\times {{10}^{-31}}}=70.66\times {{10}^{10}}\] Or \[v=8.4\times {{10}^{5}}m/s\]

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