Manipal Medical Manipal Medical Solved Paper-2012

  • question_answer
    The de-Broglie wavelength of a proton (charge\[=1.6\times {{10}^{-19}}C\].mass\[=1.6\times {{10}^{-27}}kg\]) accelerated  through  a potential difference of 1 kV is

    A)  \[600\,\overset{o}{\mathop{\text{A}}}\,\]

    B)  \[0.9\times {{10}^{-12}}m\]

    C)  \[7\,\overset{o}{\mathop{\text{A}}}\,\]            

    D)  0.9 nm

    Correct Answer: B

    Solution :

     According to de-Broglie hypothesis \[\lambda =\frac{h}{p}\] \[=\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mqV}}\] \[\therefore \]\[\lambda =\frac{6.6\times {{10}^{-34}}}{\sqrt{2\times (1.6\times {{10}^{-27}})(1.6\times {{10}^{-19}})\times 1000}}\] \[=\frac{6.6\times {{10}^{-34}}}{7.16\times {{10}^{-22}}}\] \[=0.9\times {{10}^{-12}}m\]


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