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CMC Medical CMC-Medical Ludhiana Solved Paper-2008

  • question_answer
    An electron, accelerated by a potential difference V, has de-Broglie wavelength\[\lambda .\] If the electron is accelerated by a potential difference 4V, its de-Broglie wavelength will be

    A)  \[2\lambda \]                                  

    B)  \[4\lambda \]

    C)  \[\lambda \text{/}2\]                                   

    D)  \[\lambda \text{/}4\]

    Correct Answer: C

    Solution :

                    From the relation, \[\lambda =\frac{h}{\sqrt{2\,me\,V}}\]                 ?(i) So, when V becomes 4V, then \[\lambda =\frac{h}{\sqrt{2\,me\,(4\,V)}}\] \[\Rightarrow \]               \[\lambda =\frac{h}{2\sqrt{2\,me\,V}}\] \[\Rightarrow \]               \[\lambda =\frac{\lambda }{2}\]


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