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AFMC AFMC Solved Paper-2010

  • question_answer
    A proton and an \[\alpha \]-particle are accelerated through same voltage. The ratio of their de - Broglie wavelength will be

    A)  1 : 2                                      

    B) 2\[\sqrt{2}\]: 1

    C) 2 : 1                                       

    D) \[\sqrt{2}\]: 1

    Correct Answer: B

    Solution :

    Here, \[qV=\frac{1}{2}m{{v}^{2}}\] or \[mv=\sqrt{2qmV}\] So, de-Broglie wavelength, \[\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2qmV}}\]ie, \[\lambda \propto \frac{1}{\sqrt{qm}}\] Hence, \[\frac{{{\lambda }_{p}}}{{{\lambda }_{a}}}=\sqrt{\frac{2e\times 4m}{e\times m}}=2\sqrt{2}\]


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