RAJASTHAN ­ PET Rajasthan PET Solved Paper-2003

  • question_answer
    If de-Broglie wavelength and accelerated potential difference are 6000\[\overset{o}{\mathop{\text{A}}}\,\]and 9000 V respectively, then value of \[\frac{h}{e}\] will be

    A)  \[1.9\times {{10}^{-10}}\]       

    B)  \[18\,\times \,{{10}^{-19}}\]

    C)  \[8.2\,\times \,{{10}^{-20}}\]

    D)  \[18\,\times \,{{10}^{-20}}\]

    Correct Answer: A

    Solution :

     \[\lambda =\frac{h}{\sqrt{2meV}}\Rightarrow \frac{h}{e}=\sqrt{\frac{2mv}{e}}.\lambda \] \[=\sqrt{\frac{2\times 9.1\times {{10}^{-31}}\times 9000}{1.6\times {{10}^{-10}}}}\times 6\times {{10}^{-7}}\] \[=1.9\times {{10}^{-10}}\]


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