De-broglie wavelength associated with an electron accelerated through a potential difference V is lambda. , what will b its wavelength , when accelerating potential is increased to 4V.

Here, why cant we use the formula '' lambda= 1.22/root V '' and this formula is applicable in what type of quests.?

$\lambda =\frac{12.25{A}^{o}}{\sqrt{V}}..............\left(1\right)\phantom{\rule{0ex}{0ex}}forpotentialdifference4V\phantom{\rule{0ex}{0ex}}\lambda \text{'}=\frac{12.25{A}^{o}}{\sqrt{4V}}=\frac{1}{2}\frac{12.25{A}^{o}}{\sqrt{V}}=\frac{1}{2}\lambda $

Hence wavelength become half of initial wavelength.

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