CET Karnataka Medical CET - Karnataka Medical Solved Paper-2007

  • 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 kV is

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

    B)  \[0.9\times {{10}^{-12}}m\]

    C)  \[7{{A}^{o}}\]

    D)  \[0.9\text{ }nm\]

    Correct Answer: B

    Solution :

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


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