Manipal Medical Manipal Medical Solved Paper-2002

  • question_answer
    The wavelength associated with an electron accelerated through a potential difference of 100 V is of the order of:

    A)  \[1.2\overset{o}{\mathop{\text{A}}}\,\]             

    B)  \[10.5\overset{o}{\mathop{\text{A}}}\,\]

    C)  \[100\overset{o}{\mathop{\text{A}}}\,\]    

    D)  \[1000\overset{o}{\mathop{\text{A}}}\,\]

    Correct Answer: A

    Solution :

     Here: Potentiel difference \[V=100\text{ }V\] We know that de Broglie wavelength of an electron is given by as \[\lambda =\frac{h}{\sqrt{2q\,Vm}}\] \[=\frac{6.6\times {{10}^{-34}}}{2\times (1.6\times {{10}^{-19}})\times 100\times 9.1\times {{10}^{-31}}}\] \[=1.2\times {{10}^{-10}}m=1.2\,\overset{o}{\mathop{\text{A}}}\,\]


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