Question
An electron of mass $m$ when accelerated through a potential difference $V$ has de-Broglie wavelength $\lambda $. The de-Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference will be
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Answer
(b) $\lambda = \frac{h}{{\sqrt {2mE} }}$
==> $\lambda \propto \frac{1}{{\sqrt m }}$($E =$ same)
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