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Punjab Medical Punjab - MET Solved Paper-2003

  • question_answer
    The kinetic energy of an electron is\[5\,\,eV\]. Calculate the de-Broglie wavelength associated with it (\[h=6.6\times {{10}^{-34}}\,\,Js\],\[{{m}_{e}}=9.1\times {{10}^{-31}}kg\]):

    A) \[5.47\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]                       

    B) \[10.9\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) \[2.7\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]                         

    D)  none of these

    Correct Answer: A

    Solution :

    Wavelength associated with an electron\[\lambda =\frac{h}{\sqrt{2mE}}\]                 \[=\frac{6.6\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 5\times 1.6\times {{10}^{-19}}}}\]


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