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Question

An electron of mass $$m$$ when accelerated through a potential difference $$V$$ has wavelength $$ \lambda $$. The de-Broglie wavelength associated with a proton of a mass $$M$$ through the same potential difference will be :


A
λmM
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B
λmM
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C
λMm
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D
λMm
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Solution

The correct option is B $$ \lambda \sqrt { \dfrac { m }{ M } } $$
As we knwo,
$$\lambda =\dfrac {h}{\sqrt{2mE}}\Rightarrow \lambda \propto \dfrac {1}{\sqrt m}$$   ( $$E$$= same)
so,
$$\lambda ' =\lambda \sqrt{\dfrac mM}$$

Physics
NCERT
Standard XII

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