BCECE Engineering BCECE Engineering Solved Paper-2009

  • question_answer
    The de-Broglie wavelength of a neutron at 927°C is \[\lambda \]. What will be its wavelength at 27°C?

    A) \[\frac{\lambda }{2}\]                                   

    B) \[\lambda \]

    C) \[2\lambda \]                        

    D)                         None of these

    Correct Answer: A

    Solution :

    de-Broglie wavelength \[\lambda \propto \frac{1}{\sqrt{T}}\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\frac{273+927}{273+27}}=\sqrt{\frac{1200}{300}=2}\]                 or            \[{{\lambda }_{2}}=\frac{\lambda }{2}\]


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