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AMU Medical AMU Solved Paper-2013

  • question_answer
    Energy of a photon whose de-Broglie wavelength is equal to the wavelength of an electron accelerated through a potential difference of 125 V is near to

    A)  11.5eV                

    B)  11.5keV

    C)  125eV                 

    D)  1250eV

    Correct Answer: C

    Solution :

                    Given that Potential difference V = 125 V Then,    \[\lambda =\frac{12375}{V}\overset{o}{\mathop{A}}\,\]                 \[\lambda =\frac{12375}{125}\overset{o}{\mathop{A}}\,\]                 \[\therefore \] Energy of electron in eV                 \[E=\frac{hc}{\lambda }\]                 \[E=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}\times 125\times {{10}^{-10}}}{12375}\]              = 125 eV                                          Then the energy of photon = 125 eV


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