JIPMER Jipmer Medical Solved Paper-2000

  • question_answer
    The accelerating voltage of an electron is increased to two times, its de Broglie wavelength will be:

    A)  increase 2 times             

    B)         decrease 0.5 times

    C)         increase 1.4 times          

    D)         decrease 0.7 times

    Correct Answer: D

    Solution :

    From the relation according de Broglie law \[{{\lambda }_{d}}\propto \frac{1}{\sqrt{V}}...(i)\]           [V becomes 2V (given)] \[\lambda {{}_{d}}\propto \frac{1}{\sqrt{2V}}\]                                 ?(ii) \[\frac{{{\lambda }_{d}}}{\lambda {{}_{d}}}=\sqrt{\frac{2V}{V}}\]                             hence \[\lambda {{}_{d}}=\frac{{{\lambda }_{d}}}{\sqrt{2}}\] so           \[\frac{\lambda {{}_{d}}}{{{\lambda }_{d}}}=\frac{1}{\sqrt{2}}=0.7\,\text{times}\]


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