JIPMER Jipmer Medical Solved Paper-2004

  • 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 \[\lambda =\frac{h}{\sqrt{2mE}}\] \[=\frac{6.6\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 5\times 1.6\times {{10}^{-19}}}}\] \[=5.47\times {{10}^{-10}}m=5.47\,\overset{\text{o}}{\mathop{\text{A}}}\,\]


You need to login to perform this action.
You will be redirected in 3 sec spinner